Question

a 0.02m Naoh solution and 0.01 m HCl solution are mixed in the volume ratio of 2:3 the pH of the mixed solution will be​

Answer 1

• Let volume of NaOH be 2x and HCl be 3x.

Now, they are mixed with respective concentrations:

$V_{1}S_{1} + V_{2}S_{2} = V_{3}S_{3}$

$\implies \: (0.02 \times 2x) -( 0.01 \times 3x) = (2x + 3x)S_{3}$

$\implies \: 0.04x - 0.03x = (5x)S_{3}$

$\implies \: 0.01x = (5x)S_{3}$

$\implies \: 0.01 = 5S_{3}$

$\implies \:S_{3} = 2 \times {10}^{ - 3} M$

• Since the solution is alkaline, we can say:

$\implies \:[OH^{-}]= 2 \times {10}^{ - 3} M$

$\implies \:p[OH]= - log_{10}( 2 \times {10}^{ - 3} )$

$\implies \:p[OH]= 3- log_{10}( 2 )$

$\implies \:p[OH]= 3- 0.301$

$\implies \:p[OH]= 2.699$

$\implies \:p[H]= 14 - 2.699$

$\implies \:p[H]= 11.301$

So, pH of mixture is 11.301