Question

The number of $H^{+}$ present in one mL of solution is $6.022\times {10}^x$ whose $pH$ is $13$. Find out the value of x.

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

Answer: (B)

7

As we know, $pH=-log\left[H^{+}\right]$

For $pH=13$,

$\left[H^{+}\right]={10}^{-pH}={10}^{-13}M$ .............(1)

So number of $\left[H^{+}\right]$ in 1 ml,

$\left[H^{+}\right]={10}^{-13}=\frac{6.023\times {10}^x}{N_A\times 0.001}$ ...........(2)

On equating equation 1 and 2, we get-

So $x=7$.