Question

Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M.

Answer 1

Answer:

For 0.1 M sodium acetate solution

if concentration of acid is 0.0025 then pH will  6.075

if concentration of acid is 0.005 then pH will  5.775

if concentration of acid is 0.01 then pH will  5.475

if concentration of acid is 0.05 then pH will  4.775

For 0.01 M sodium acetate solution

if concentration of acid is 0.0025 then pH will  5.075

if concentration of acid is 0.005 then pH will  4.775

if concentration of acid is 0.01 then pH will  4.475

if concentration of acid is 0.05 then pH will  3.775

Explanation:

to calculate the pH of a buffer solution we use the following formula

pH = pKa + log [B]/[A] ------------- eq (1)

[B] = concentration of base

[A] = concentration of acid

Given data

[B] = 0.1 M , 0.01M

[A] = 0.0025 M , 0.005 M, 0.01 M, 0.05 M

pKa value for sodium acetate is 4.75

1. First we will calculate the pH values for 0.1 M acetate solution.

If the concentration of acid is 0.0025, then:

[B] = 0.1 M

[A] = 0.0025 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.0025]

pH = 4.75 + log [40]

pH = 4.475 + 1.6

pH = 6.075

If the concentration of acid is 0.005 M, then:

[B] = 0.1 M

[A] = 0.005 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.005]

pH = 4.75 + log [20]

pH = 4.475 + 1.3

pH = 5.775

If the concentration of acid is 0.01, then:

[B] = 0.1 M

[A] = 0.01 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.01]

pH = 4.75 + log [10]

pH = 4.475 + 1

pH = 5.475

If the concentration of acid is 0.05, then:

[B] = 0.1 M

[A] = 0.05 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.1]/[0.05]

pH = 4.75 + log [2]

pH = 4.475 + 0.3

pH = 4.775

2. Now we will calculate the pH values for 0.01 M acetate solution.

If the concentration of acid is 0.0025, then:

[B] = 0.01 M

[A] = 0.0025 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.0025]

pH = 4.75 + log [4]

pH = 4.475 + 0.6

pH = 5.075

If the concentration of acid is 0.005 M, then:

[B] = 0.01 M

[A] = 0.005 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.005]

pH = 4.75 + log [2]

pH = 4.475 + 0.3

pH = 4.775

If the concentration of acid is 0.01 M, then:

[B] = 0.01 M

[A] = 0.01 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.01]

pH = 4.75 + log [1]

pH = 4.475 + 0

pH = 4.475

If the concentration of acid is 0.05 M, then:

[B] = 0.01 M

[A] = 0.05 M

put these values in eq 1. which is:

pH = pKa + log [B]/[A]

pH = 4.75 + log [0.01]/[0.05]

pH = 4.75 + log [0.2]

pH = 4.475 + (-0.7)

pH = 4.475 - 0.7

pH = 3.775

Answer 2

Answer:

a) pH = 4.71

b) pH = 4.704

c) pH = 4.57

d) No buffer here, the pH will be between 2-3

Explanation:

Applying Henderson Hasselbach equation:

pH = pKa + log([A]/[HA])

a) For 0.0025 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.0025 M of acid:

[A] = 0.05 - 0.0025 = 0.0475 M

[HA] = 0.05 + 0.0025 = 0.0525 M

$pH=4.75+log(\frac{0.0475}{0.0525} )=4.71$

b) For 0.005 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.005 M of acid:

[A] = 0.05 - 0.005 = 0.0495 M

[HA] = 0.05 + 0.005 = 0.055 M

$pH=4.75+log(\frac{0.0495}{0.055} )=4.704$

c) For 0.01 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.01 M of acid:

[A] = 0.05 - 0.01 = 0.04 M

[HA] = 0.05 + 0.01 = 0.06 M

$pH=4.75+log(\frac{0.04}{0.06} )=4.57$

d) For 0.05 M:

[A] = 0.1/2 = 0.05 M

[HA] = 0.05 M

After add 0.05 M of acid:

[A] = 0.05 - 0.05 = 0

[HA] = 0.05 + 0.05 = 0.1 M

No buffer here, the pH will be between 2-3