Question

A 50 mL solution of pH=1 is mixed with 50mL soltion of pH =2 .The pH of new mixture will be nearly (use log 5.5= 0.74)

Answer 1

The new pH of mixture is 1.26

Answer 2

Answer : The pH of resulting solution is, 1.26

Solution :

First we have to calculate the concentration from the pH = 1 and pH = 2

concentration at pH = 1

$pH_1=-\log [H_1^+]\\\\1=-\log [H_1^+]$

$[H_1^+]=0.1$

concentration at pH = 2

$pH_2=-\log [H_2^+]\\\\2=-\log [H_2^+]$

$[H_2^+]=0.01$

$[H_1^+]$ and $[H_2^+]$ are the concentrations.

Now we have to calculate the concentration of resulting solution.

Formula used :

$C_1V_1+C_2V_2=C_3V_3$

where,

$C_1$ and $C_2$ are the concentrations

$C_3$ is the concentration of resulting solution

$V_1$ and $V_2$ are the volume = 50 ml = 0.05 L

$V_3$ is the volume of resulting solution = 50 + 50 = 100 ml  = 0.1 L

Now put all the given values in the above formula, we get the concentration of resulting solution.

$(0.1)\times 0.05+(0.01)\times 0.05=C_3\times 0.1$

$C_3=0.055$

The concentration of the resulting solution = $0.055$

Now we have to calculate the pH of the resulting solution.

$pH_3=-\log [H_1^+]\\\\pH_3=-\log (0.055)=1.26$

Therefore, the pH of resulting solution is, 1.26