Question

How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.

Answer 1

Here is the full question

How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.  Ka of acetic acid is 1.7×10-5.

Answer:

0.3396 moles

Explanation:

pH = pKA + log$\frac{[acetate]}{[aceticacid]}$

[acetic acid] = 0.1 M

pH = 5.0

$K_a$ = $1.7*10^{-5$

5.0 = -log ($1.7*10^{-5$) + log $\frac{[acetate]}{0.1}$

5.0 = 4.77 + log [acetate] + 1

5.0 = 5.77 + log [acetate]

- log [acetate] = 5.77 - 5.0

- log [acetate]  = 0.77

log [acetate]  = -0.77

[acetate]  = log⁻¹ (-0.77)

[acetate]  =  0.1698

∴ for 2L = 2 × 0.1698

= 0.3396 moles

Answer 2

The sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles

Given:

The concentration of acetic acid = 0.10 M

pH = 5.0  

$K_a$ = $1.7\times10^{-5$

Solution:

According to Hasselbalch equation:

$pH=pKa + log\frac{[A^-]}{[HA]}\\\\or\\\\pH = pKa + log\frac{(acetate)}{(acetic acid)}$

5.0 = -log $(1.7\times10^{-5})$ + log $\frac{(acetate)}{(0.1)}$

5.0 = 4.77 + log [acetate] + 1

5.0 = 5.77 + log [acetate]

- log [acetate] = 5.77 - 5.0

- log [acetate]  = 0.77

log [acetate]  = -0.77

[acetate]  = log⁻¹ (-0.77)

[acetate]  =  0.1698

∴ for 2L = 2 × 0.1698

= 0.3396 moles

Thus, the sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles

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