Question

Two buffers (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffer ? ($K_{HA}$ = 1.0 × $10^{-5}$)

Answer 1

In presence of HA:

$pH\quad =\quad -logKa\quad +\quad log\left[ \frac { [Salt] }{ [Acid] } \right]$

From the given

pH of buffer = 4

$4\quad =\quad -log(1.0\quad \times \quad { 10 }^{ -5 })\quad +\quad log\left[ \frac { [Salt] }{ 0.5 } \right]$

Therefore, $log\left[ \frac { [Salt] }{ 0.5 } \right] \quad =\quad -1$

Or

$[Salt]\quad =\quad 0.1\quad \times \quad 0.5\quad =\quad 0.05\quad M$

In presence of NaA:

$pH\quad =\quad 6$

$6\quad =\quad -\log { \left( 1.0\quad \times \quad { 10 }^{ -5 } \right)  } \quad +\quad log\left[ \frac { [Salt] }{ 0.5 }\right]$

Therefore,$\quad log\left[ \frac { [Salt] }{ 0.5 }\right] \quad =\quad -1$

Therfore,$\quad [Salt]\quad =\quad 10\quad \times \quad 0.5\quad =\quad 5\quad M$

Now, the two buffer

(i) NaA = 0.05 M and HA = 0.5 M

(ii) NaA = 5 M and HA = 0.5 M)

These are mixed in equal proportion.

New concentration of NaA in mixed buffer $=\quad \frac { (0.05\quad \times \quad V\quad +\quad 5\quad \times \quad V) }{ 2V } \quad =\quad \frac { 5.05 }{ 2M }$

New concentration of HA in mixed buffer $=\quad \frac { (0.5\quad \times \quad V\quad +\quad 5\quad \times \quad V) }{ 2V } \quad =\quad 0.05$

Therefore, $pH\quad =\quad -log1.0\quad \times \quad { 10 }^{ -5 }\quad +\quad log\left[ \frac { \frac { 5.05 }{ 2M } }{ 0.05 } \right] \quad =\quad 5\quad +\quad 0.7033\quad =\quad 5.7033$

Answer 2

☺☺hey mate


pH= 5.7033.........______Answer....☺✌☺