Question

When equal volume of ph4 and ph 6 are mixed then ph of resulting solution

Answer 1

The formula you use here will be
N1V1+N1V2=NV
SINCE PH =4
H+=10^-4
PH=6
H+=10^-6

10^-4*V+10^-6*V =N2V
take v common and as you solve further you'll find the answer for N which will be the PH resulting in answer

Answer 2

Answer:

The pH of the final solution is 4.29.

Explanation:

Solution with pH = 4

$pH=-\log[H^+]$

$4=-\log[H^+]$

$[H^+]=10^{-4} mol/L$

Moles of $H^+$ in V volume:

$=V\times 10^{-4} mol/L$

Solution with pH = 6

$pH=-\log[H^+]'$

$6=-\log[H^+]'$

$[H^+]'=10^{-6} mol/L$

Moles of $H^+'$ in V volume:

$=V\times 10^{-6} mol/L$

Let V volume of each solution is mixed:

Total volume of the final solution = 2V

Concentration of $H^+$ in the final solution:

$[H^+]=\frac{V\times 10^{-4} mol/L+V\times 10^{-6} mol/L}{2V}$

$=0.0000505 mol/L$

The pH of the final solution:

$pH=-\log[0.0000505]=4.29$