how to solve buffer solution problems

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Chemistry library

Course: chemistry library   >   unit 14.

• Introduction to buffers
• Properties of buffers
• Henderson–Hasselbalch equation
• Common ion effect and buffers

Buffer solution pH calculations

• Methods for preparing buffers
• pH and pKa relationship for buffers
• Buffer capacity

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Video transcript

  [base] pH = pK a + log –––––   [acid]

pH = pK a + log [base / acid]

The K a of acetic acid is 1.77 x 10¯ 5

pK a = −log K a = −log 1.77 x 10¯ 5 = 4.752

  1.00 pH = 4.752 + log –––––   1.00 Since the log of 1 is zero, we have pH = 4.752

HAc ⇌ H + + Ac¯

  1.00 pH = 4.752 + log –––––   0.800 x = 4.752 + 0.097 = 4.849

pH = pK a + log [base / acid] x = 4.752 + log (0.800 / 1.00) x = 4.752 − 0.097 = 4.655

pH = pK a + log [base / acid] pH = 3.752 + log [0.500 / 0.700] pH = 3.752 + (−0.146) pH = 3.606

HCOOH ---> (0.700 mol/L) (0.500 L) = 0.350 mol HCOONa ---> (0.500 mol/L) (0.500 L) = 0.250 mol

NaOH ---> (1.00 mol/L) (0.0500 L) = 0.0500 mol

HCOOH ---> 0.350 mol − 0.0500 mol = 0.300 mol HCOONa ---> 0.250 mol + 0.0500 mol = 0.300 mol

pH = 3.752 + log [0.300 / 0.300] pH = 3.752 + log 1 pH = 3.752

(a) a strong acid (or base) is added to an already-buffered solution (b) a strong acid (or base) is added to a solution with just a weak base (or weak acid) in it, creating a buffer.

CH 3 NH 2 + H + ---> CH 3 NH 3 +

0.02 mol CH 3 NH 2 0.08 mol CH 3 NH 3 +

First, however, we will use the K b to get the pK a . pK a + pK b = 14 pK b = −log 5 x 10¯ 4 = 3.30103 pK a = 14 − 3.30103 = 10.69897

pH = 10.69897 + log (0.02 / 0.08) pH = 10.69897 + (−0.60206) pH = 10.09691

[H + ] = 10¯ pH = 10¯ 10.09691 = 8 x 10¯ 11 M

(0.200 mol/L) (0.0250 L) = 0.00500 mol of acetic acid (0.100 mol/L) (0.0350 L) = 0.00350 mol of NaOH

acetic acid ---> 0.00500 mol − 0.00350 mol = 0.00150 mol The acetic acid that reacts with the NaOH produces sodium acetate. In the solution will be 0.00350 mol of acetate anion (we may ignore the sodium ion. It plays no role in the pH.)

pH = 4.752 + log (0.00350 / 0.00150) pH = 4.752 + 0.368 pH = 5.120

(0.180 mol/L) (0.0500 L) = 0.00900 mol of ammonia (0.360 mol/L) (0.0050 L) = 0.00180 mol of HBr

ammonia ---> 0.00900 mol − 0.00180 mol = 0.00720 mol

NH 3 + HBr ---> NH 4 Br We can see that the ammonium ion is produced in a 1:1 molar ratio with each reactant. Since HBr is the limiting reagent, we determine that 0.00180 mole of ammonium ion will be produced. (We may ignore the bromide. It plays no role in the pH.)

pH = 9.248 + log (0.00720 / 0.00180) pH = 9.248 + 0.602 pH = 9.850

K a K b = K w (K a ) (1.77 x 10¯ 5 ) = 1.00 x 10¯ 14 K a = 5.64972 x 10¯ 10 pK a = −log 5.64972 x 10¯ 10 = 9.248

pH = pK a + log [0.25 / 0.35] The molarity of the acid (the ammonium chloride is the acid) was arrived at by dividing 0.35 mol by 1.0 L.

K w = K a K b 1.00 x 10¯ 14 = (K a ) (1.77 x 10¯ 5 ) K a = 5.65 x 10¯ 10 pK a = −log 5.65 x 10¯ 10 = 9.248 By the way, providing one constant (the K b in this example) while the solution requires a different, but related, constant (the pK a in this example) is very common in problems of this type.

pH = 9.248 + log [0.25 / 0.35] pH = 9.248 + (−0.146) = 9.10

"0.500 L of an acetic acid buffer (0.800 M total) at maximum buffering capacity" There is a lot of information in that sentence. First, since it is an "acetic acid buffer" we know that both acetic acid CH 3 COOH (or HAc) and its conjugate base acetate ion CH 3 COO¯ (or Ac¯) are present. Maximum buffering capacity means that the acid and its conjugate base are in a 1:1 molar ratio.

0.800 M (total) means 0.800 mol (total) per liter We have 0.500 L present, which means 0.400 mol (total) present Maximum buffering capacity requires that HAc and Ac¯ be present in a 1:1 molar ratio. Since 0.400 mol (total) is present, that means this is present in 0.500 L: moles HAc = 0.200 mol moles Ac¯ = 0.200 mol

moles HAc ---> 0.200 + 0.100 = 0.300 mol moles acetate ---> 0.200 − 0.100 = 0.100 mol

pH = 4.752 + log (0.100 / 0.300) pH = 4.752 + (−0.477) pH = 4.275

HAc ---> 0.400 mol salt ---> 0.400 mol

HAc ---> 0.400 mol salt ---> 0.400 + 0.100 = 0.500 mol

pH = 4.752 + log (0.500 / 0.400) pH = 4.752 + 0.097 pH = 4.849

(a) Calculate the molecular weight of the acid. (b) Calculate the K a of the acid. (c) Calculate the pOH of the original solution. (d) Calculate the pH after 40.0 mL of NaOH solution was added. (e) Calculate the percent ionization of the acid in the original solution. (f) Calculate the pH at the equivalence point.

Note that as a result of the titration TWO pieces of data emerge: (i) After we add 16 mL of the NaOH, the pH = 4.250. (ii) After we add 80 mL of the NaOH, the equivalence point is reached.

HA + OH¯ ---> H 2 O + A¯ This is commonly called a neutralization reaction.

moles = (0.500 mol/L) (0.0800 L) = 0.0400 mol

(i) the moles of strong base added (here NaOH) equals the original number of moles of acid (here HA) (ii) the titrant (here NaOH) is "chemically equivalent" to the analyte (here HA) (iii) enough NaOH has been added to "neutralize" the original HA present Based on the 1:1 molar ratio between HA and NaOH (see chemical equation above), we conclude that 0.0400 mol of acid was originally present. In other words, at the equivalence point, the moles of NaOH added equals the moles of HA present before reaction.

4.92 g / 0.0400 mol = 123 g/mol

moles = (0.500 mol/L) (0.0160 L) = 0.00800 mol

0.0400 mol − 0.00800 mol = 0.0320 mol of HA remaining. Please note that 0.0080 mol of A¯ is produced.

  [base] pH = pK a + log –––––   [acid]   0.0080 4.250 = pK a + log –––––   0.0320 Note the direct use of moles rather than molarities. 4.250 = pK a + (−0.602) pK a = 4.852 K a = 1.406 x 10¯ 5 Which, to three sig figs, is 1.41 x 10¯ 5

  [H 3 O + ] [A¯] K a = ––––––––––   [HA]   (5.623 x 10¯ 5 ) (0.00800 + 5.623 x 10¯ 5 ) K a = –––––––––––––––––––––––––––––––   0.0320 − 5.623 x 10¯ 5 Notice that this set up includes the amount that the HA would go down and the amount that the A¯ would go up when producing the H 3 O + in the solution. The Henderson-Hasselbalch Equation ignores amount, making the H-H an approximation. The [H 3 O + ] came from the pH of 4.250.   (5.623 x 10¯ 5 ) (0.00805623) K a = ––––––––––––––––––––––   0.03194377 Ka = 1.418 x 10¯ 5 Which, to three sig figs, is 1.42 x 10¯ 5

HA + H 2 O ⇌ H 3 O + + A¯   [H 3 O + ] [A¯] K a = ––––––––––   [HA]   (x) (x) 1.406 x 10¯ 5 = –––––––––   0.0800 − x Note that I used a slightly less-rounded off value for the K a . Ignoring the 'subtract x' leads to: x = (1.406 x 10¯ 5 ) (0.0800) x = 0.001060566 M pH = 2.974 pOH = 11.026

Percent ionization = the hydrogen ion concentration divided by the original acid concentration times 100 (0.0010606 M / 0.0800 M) * 100 = 1.32575% to three sig figs, 1.32%

A¯ + H 2 O ⇌ HA + OH¯   [HA] [OH¯] K b = ––––––––––   [A¯]

K a K b = K w (1.406 x 10¯ 5 ) (K b ) = 1.00 x 10¯ 14 K b = 7.11238 x 10¯ 10 [A¯] = 0.0800 mol / 0.580 L = 0.137931 M Note use of the combined volumes (500 mL of weak acid mixed with 80 mL of NaOH solution.

  (x) (x) 7.11238 x 10¯ 10 = ––––––––   0.137931 x = 9.90463 x 10¯ 6 M

pOH = −log 9.90463 x 10¯ 6 = 5.004 pH = 8.996

(1.80 mol/L) (0.00120 L) = 0.00216 mol

(0.850 mol/L) (0.02100 L) = 0.01785 mol

0.01785 mol − 0.00216 mol = 0.01569 mol By the way, 0.00216 mol of ammonium ion was produced by the H + reacting with the NH 3 .

pK a + pK b = pK w pK b = −log 1.77 x 10¯ 5 = 4.752 pK a + 4.752 = 14 pK a = 9.248

  [base] pH = pK a + log –––––   [acid]   0.01569 pH = 9.248 + log –––––––   0.00216 pH = 9.248 + 0.861 = 10.109

  [base] pH = pK a + log –––––   [acid]   0.200 pH = 9.248 + log –––––   0.150 pH = 9.248 + 0.125 = 9.373 Note that I used the pK a of ammonium ion.

First, you need to know the volume of buffer into which that 0.02 mol of NaOH has been added. This is because you need to know the actual moles of NH 3 and NH 4 + in the original solution, not just their concentration. So, for example, assume you have 500 mL of that original buffer. In that solution you have: (0.5 L) (0.20 mol/L) = 0.10 mol NH 3 (0.5 L) (0.15 mol/L) = 0.075 mol NH 4 +

moles NH 3 = 0.10 + 0.02 = 0.12 mol NH 3 moles NH 4 + = 0.075 − 0.02 = 0.055 mol NH 4 +

(1.00 L) (0.20 mol/L) = 0.20 mol NH 3 (1.00 L) (0.15 mol/L) = 0.15 mol NH 4 + moles NH 3 = 0.20 mol + 0.02 mol = 0.22 mol moles NH 4 + = 0.15 mol − 0.02 mol = 0.13 mol pH = 9.25 + log (0.22 / 0.13) = 9.48

MV = grams / molar mass (M) (0.250 L) = 7.45 g / 53.4916 g/mol M = 0.557097 mol/L

M 1 V 1 = M 2 V 2 (2.32 mol/L) (60.00 mL) = (x) (250.0 mL) x = 0.5568 M

  0.557097 pH = 9.248 + log –––––––   0.5568 pH = 9.248 + 0.0002316 pH = 9.248

(a) The HCl protonates some, but not all, of the NH 3 . You still have a buffer in that case. In this situation, the Henderson-Hasselbalch Equation is used. (b) There is exactly enough HCl to neutralize all of the NH 3 , leaving only NH 4 Cl in solution. That is a solution of a salt of a weak base, it is not a buffer. Do a K a caculation using the K a of NH 4 + . (c) There is excess HCl left after all the NH 3 has been protonated. In that case, you ignore all the NH 4 Cl that is in solution and treat the solution as having only a strong acid in it. The solution is not a buffer, nor is it a weak acid calculation. You are simply calculating the pH of a solution of a strong acid.

moles = MV = (0.150 mol/L) (0.500 L) = 0.0750 mol

NH 3 ---> 0.250 mol NH 4 Cl ---> 0.225 mol

NH 3 ---> 0.250 mol − 0.0750 mol = 0.175 mol NH 4 Cl ---> 0.225 mol + 0.0750 mol = 0.300 mol Note that HCl reacts with NH 3 to form NH 4 Cl in a 1:1:1 molar ratio.

pH = pK a + log ([base] / [acid]) pH = 9.248 + log (0.175 / 0.300) pH = 9.248 + (−0.234) pH = 9.014 Note that I used the pK a of ammonium, not the pK b of ammonia.

pH = 9.248 + log (0.250 / 0.225) pH = 9.248 + 0.046 pH = 9.294 Adding the HCl (an acid) has changed the pH of the buffer in the acidic direction, from 9.294 to 9.014.

moles = MV = (0.500 mol/L) (0.500 L) = 0.250 mol

NH 3 ---> 0.250 mol − 0.250 mol = 0 mol NH 4 Cl ---> 0.225 mol + 0.250 mol = 0.475 mol

NH 4 + + H 2 O ⇌ H 3 O + + NH 3

  [NH 3 ] [H 3 O + ] K a = ––––––––––––   [NH 4 + ]

  (x) (x) 5.64937 x 10¯ 10 = ––––––––   0.158333 x = 0.0000094577 M pH = −log 0.0000094577 = 5.024 0.158333 M comes from 0.475 mol divided by 3.00 L

moles = MV = (0.650 mol/L) (0.500 L) = 0.325 mol

NH 3 ---> 0.250 mol − 0.325 mol = −0.075 mol The NH 3 is completely consumed and there is 0.075 mol of HCl left over.

0.075 mol / 3.00 L = 0.025 M

pH = −log 0.025 = 1.60 Remember, in the presence of a strong acid (HCl), the weak acid (NH 4 Cl) plays no role in determining the pH.

(a) The NaOH deprotonates some, but not all, of the NH 4 Cl. You still have a buffer in that case. In this situation, the Henderson-Hasselbalch Equation is used. (b) There is exactly enough NaOH to neutralize all of the NH 4 Cl, leaving only NH 3 in solution. That is a solution of a weak base, it is not a buffer. Do a K b caculation using the K b of NH 3 . (c) There is excess NaOH left after all the NH 3 has been deprotonated. In that case, you ignore all the NH 3 that is in solution and treat the solution as having only a strong base in it. The solution is not a buffer, nor is it a weak acid calculation. You are simply calculating the pH of a solution of a strong base.

NH 3 ---> 0.250 mol + 0.0750 mol = 0.325 mol NH 4 Cl ---> 0.225 mol − 0.0750 mol = 0.150 mol Note that NaOH reacts with NH 4 Cl to form NH 3 in a 1:1:1 molar ratio.

pH = pK a + log ([base] / [acid]) pH = 9.248 + log (0.325 / 0.150) pH = 9.248 + (0.336) pH = 9.584 Note that I used the pK a of the ammonium ion, not the pK b of ammonia.

pH = 9.248 + log (0.250 / 0.225) pH = 9.248 + 0.046 pH = 9.294 Adding the NaOH (a base) has changed the pH of the buffer in the basic direction, from 9.294 to 9.584.

moles = MV = (0.450 mol/L) (0.500 L) = 0.225 mol

NH 3 ---> 0.250 mol + 0.225 mol = 0.475 mol NH 4 Cl ---> 0.225 mol − 0.225 mol = 0 mol

NH 3 + H 2 O ⇌ NH 4 + + OH¯

  [NH 4 + ] [OH¯] K b = ––––––––––––   [NH 3 ]

  (x) (x) 1.77 x 10¯ 5 = ––––––––   0.1583333 x = 0.00167407 M pOH = −log 0.00167407 = 2.776 pH = 14 − 2.776 = 11.224 0.1583333 M comes from 0.475 mol divided by 3.00 L

moles = MV = (0.650 mol/L) (0.500 L) = 0.350 mol

NH 4 Cl ---> 0.225 mol − 0.350 mol = −0.125 mol There is 0.125 mol of NaOH left over.

0.125 mol / 3.00 L = 0.04166667 M

pOH = −log 0.04166667 = 1.380 pH = 14 − 1.380 = 12.620 Remember, in the presence of a strong base (NaOH), the weak base (NH 3 ) plays no role in determining the pH.

(0.664 mol/L) (0.975 L) = 0.6474 mol

pH = pK a + log [base / acid] 6.420 = 4.752 + log [x / (0.6474 − x)] The x is the moles of acetate that must be present and the 0.6474 − x is the amount of acetic acid.

log [x / (0.6474 − x)] = 1.668 [x / (0.6474 − x)] = 46.5586 x = 30.142 − 46.5586x 47.5586x = 30.142 x = 0.63379 mol Acetate and KOH are in a 1:1 stoichiometric ratio, so this is the required number of moles of KOH.

0.63379 mol / 2.50 mol/L = 0.253516 L 254 mL seems like a reasonable answer

pH = 4.752 + log (0.63379 / 0.01361) = 6.420

Chemistry Steps

General Chemistry

Acids and bases.

This is a summary practice problem set on buffer solutions aimed to help identify buffers, calculating the pH of a buffer solution prepared from a weak acid and its conjugate base or vice versa.

The links to the corresponding topics are given below:

• Buffer Solutions
• The Henderson–Hasselbalch Equation
• The pH of a Buffer Solution
• Preparing a Buffer with a Specific pH
• The Common Ion Effect
• The pH and p K a  Relationship

Which of the following solutions can be classified as buffer solutions?

(a) NaBr + HBr, (b) NaHSO 4 + H 2 SO 4 , (c) HCl + HOCl (d) Na 2 HPO 4 + NaH 2 PO 4 , (e) CH 3 CH 2 NH 2 + CH 3 CH 2 NH 3 + (f) NaNO 2 + HNO 2 , (g) KCN + HCN, (h) Na 2 SO 4 + NaHSO 4 , (i) NH 3 + NH 4 ClO 3 , (j) CH 3 CO 2 H + NaOH.

Addition of which salt will suppress the ionization of HOCl?

A solution of ammonia and ammonium chloride is a common buffer system. Write the corresponding reactions to show how it resists a pH change when an acid or a base is added to it.

Calculate the pH of the buffer solution consisting of 0.75 M NH 3 and 0.95 M NH 4 Cl. K b (NH 3 ) = 1.8 x 10 -5 .

Calculate the pH of the buffer solution which is 1.4 M CH 3 COONa and 1.8 M CH 3 COOH. K a  (CH 3 CO 2 H) = 1.7 x 10 -5

Calculate the pH of the buffer solution that is 0.70 M NaCN and 0.55 M HCN. K a  (HCN) = 4.9 x 10 -10

Calculate the pH of the buffer solution that is 0.60 M propionic acid (CH 3 CH 2 CO 2 H) and 0.75 M sodium propionate. K a  (CH 3 CH 2 CO 2 H) = 1.32 x 10 -5

What is the pH of a buffer solution that is 0.85 M pyridine (C 5 H 5 N) and 1.3 M pyridinium chloride (C 5 H 5 NHCl)? K b (C 5 H 5 N) = 1.7 x 10 -9 16.82

Calculate the pH of a solution that is 0.8 M HF and 2.0 M NaF. K a (HF) = 6.6 x 10 -4

Calculate the ratio of NaNO 2 to HNO 2 required to create a buffer with pH = 4.00. K a (HNO 2 = 7.2 x 10 -4 )

Assuming no volume change, how many grams of sodium benzoate (C 7 H 5 O 2 Na) needs to be added to 250.0 mL of a 0.25 M benzoic acid (C 7 H 5 O 2 H) solution to prepare a buffer with a pH of 4.60? K a = 6.46 x 10 -5

How many mL of 0.60 M HF and 0.70 M NaF must be mixed to prepare 1.00 L of a buffer solution at pH of 4.2? p K a HF = 3.8

What is the concentration of C 6 H 5 NH 3 Cl in the buffer solution with a pH of 4.3 containing 0.50 M C 6 H 5 NH 2 ? K b (C 6 H 5 NH 2 ) 3.8 x 10 -10

Which of the following pairs is the best choice to prepare a buffer with pH = 3.2 ?

In what mass ratio would you add the components to prepare the buffer.

a) CH 3 COOH and CH 3 COONa b) HNO 2 and KNO 2 c) NH 3 and NH 4 Cl d) C 5 H 5 N and C 5 H 5 NHCl.

What would be the pH of a buffer solution composed of 0.60 M NH 3 /0.45 M NH 4 Cl after 0.010 mole of gaseous HCl is added to a 400.0 mL solution?

Calculate the pH after 20.0 g NaOH( s ) is added to 1.0 L of a buffer solution containing 1.80 M acetic acid and 1.50 M sodium acetate at pH 4.0.

• The pH and p K a  Relationship
• Strong Acid–Strong Base Titrations
• Titration of a Weak Acid by a Strong Base
• Titration of a Weak Base by a Strong Acid
• Titration of Polyprotic Acids
• Buffer Solutions Practice Problems
• K sp  and Molar Solubility
• The Effect of a Common Ion on Solubility
• The Effect of pH on Solubility
• Will a Precipitate Form?  K sp  and  Q
• K sp  and Molar Solubility Practice Problems

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