Question

A hydrogen electrode 'X' was placed in a buffer solution. 'P' [containing potassium formate (2 moles) and formic acid (1 mole)] and another hydrogen electrode 'Y' was, placed in a buffer solution 'Q' [(containing potassium formate (1 mole) and formic acid (2 moles)], The reduction potential values for the cells X and Y are found to be -0.50 V and -0.09 V respectively w.r.t. standard hydrogen electrode.
Now, if 0.5 moles of HCl are added in each of the buffer P & Q, then pH of buffer 'P' and 'Q' respectively are: (Assume partial pressure of $H_2$ gas' to be 1 atm and temperature 298 K, log5=0.7)

• A. 5.3, 5.0
• B. 4.3, 4.0
• C. 5.0, 5.7
• D. 5.0, 4.3

Answer 1

Answer: (D)

5.0, 4.3

$H^{+}\left(aq\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right)$

$E_{red}=E_{red}^o+\frac{0.059}{1}log\left[H^{+}\right]$

$-E_{red}=-E_{red}^o-0.0591\times log\left[H^{+}\right]$

$-E_{red}=0+0.0591\times pH$

$-\frac{E_{red}}{0.059}=pH=p{K}_a+log\frac{\left[HCO{O}^{-}\right]}{\left[HCOOH\right]}$

For electrode 'X',

$\frac{-(-0.50)}{0.059}=p{K}_a+log\frac{2}{1}$ ....(1)

$\frac{-(-0.090)}{0.059}=p{K}_a-log\frac{2}{1}$

Adding (1) & (2)

$\frac{0.590}{0.059}=2p{K}_a$

$\therefore p_{K_a}=\frac{0.59}{0.118}=5$

Now, for buffer (P) (after added 0.5 moles of HCl)

$pH=5+log\frac{1.5}{1.5}=5$

And for buffer 'Q', (after adding HCl)

$p{H}^1=5+log\frac{0.5}{2.5}=4.3$.