Question

How much volume of 0.3 MNH₄OH should be added to 60 mL of 0.1 MNH₄Cl so that pH of the buffer solution is 10? Given that pK_b for NH₄OH = 4.75.

Answer 1

6. According to Henderson - hassel blach equation for base buffer

P^OH - P^kb + log \(\frac{[salt]}{[base]}\) ----- (1)

we have given,

P^H of buffer solution = 10

∴ P^OH = 14 - P^H

 = 14 - 10

P^OH = 4

Let say volume of Hn₄OH = VmL

Then number of moles of NH₄OH = V x 0.3 milimol

Total volume of buffer solution = (V + 60) ml

Number of moles of NH₄CL = 60 x 0.1 = 6 milimol

Putting the values in equation (1) -- we got.

taking anti log on both side---

0. 178 = \(\frac{6}{V\times0.3}\)

\(\Rightarrow{}\)v = \(\frac{6}{0.178\times0.3}\)

\(\Rightarrow{}\) v = 112.4 ml

Hence, 112.4 ml of 0.3 M NH₄OH should be added.

7. According to Henderson - Hassel blach equation for acid buffer --

P^H = P^KA + log \(\frac{[salt]}{[acid]}\) -----(II)

we have given,

concertrational of CH₃COOH = 0.01 M

concertrational of CH₃COOHa = 0.4 M

volume of CH₃COOH = V ML

Let see, Volume of CH₃COOH = V ML

Then, Total volume of buffer solution = (V + 25) ML

P^H of buffer solution = 4.91

P^KA of acid = 4.76

After mixing --- 

concertrational of CH₃COOH = \(\frac{V\times0.01}{V+25}\) M

concertrational of CH₃COOHa = \(\frac{10}{V+25}\) M

Putting the values in equation (II) we got --

4.91 = 4.76 + log \(\frac{(\frac{10}{V+25})}{(\frac{V\times0.01}{V+25})}\) 

\(\Rightarrow\) 0.15 = log \(\frac{10}{V\times0.01}\)

taking antilog on both side --- we got ---

\(\Rightarrow\) 1.412537 = \(\frac{10}{V\times0.01}\)

\(\Rightarrow\) V = 707.946 ML

\(\Rightarrow\) V = 707.9 ML