Question

The decrease in % in $\left[H^{+}\right]$ when $pH$ increases by $0.1$ will be:

• A. 10
• B. 30
• C. 50
• D. 20

Answer 1

Answer: (D)

20

$pH=-log\left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-pH}$

So, % change in hydrogen ion $=\frac{({10}^{-\left(pH\right)}-{10}^{-(pH+0.1)})\times 100}{{10}^{-pH}}=(1-{10}^{-0.1})\times 100\approx 20$