Question

4. Equal volumes of three monobasic acids of pH
3,4,5 are mixed in a vessel. The pH of the resultant
solution will be (log 3.7 = 0.57)
(1) 3.91
(2) 5.27
(3) 6.81
(4) 3.43​

Answer 1

Given:

Equal volumes of three monobasic acids of pH 3,4,5 are mixed in a vessel.

To find:

pH of solution ?

Calculation:

So, the hydrogen ion concentrations from the constituent acids are :

$1) \: pH = 3 \\ \implies \: [{H}^{ + } ] = {10}^{ - 3} \: M$

$2) \: pH = 4 \\ \implies \: [{H}^{ + } ] = {10}^{ - 4} \: M$

$3) \: pH = 5 \\ \implies \: [{H}^{ + } ] = {10}^{ - 5} \: M$

Now, net concentration in solution:

$\: [{H}^{ + } ]_{net} = \dfrac{ {10}^{ - 3} + {10}^{ - 4} + {10}^{ - 5} }{3}$

$\implies \: [{H}^{ + } ]_{net} = \dfrac{0.00111}{3}$

$\implies \: [{H}^{ + } ]_{net} = 0.00037$

$\implies \: [{H}^{ + } ]_{net} = 3.7 \times {10}^{ - 4}$

Now, pH will be :

$\implies \: pH = - log \bigg([{H}^{ + } ]_{net} \bigg)$

$\implies \: pH = - log \bigg( 3.7 \times {10}^{ - 4} \bigg)$

$\implies \: pH =4 - log \bigg( 3.7 \bigg)$

$\implies \: pH =4 -0.57$

$\implies \: pH =3.43$

So, pH of solution will be 3.43 .