Question

To a 100 ml of an aqueous HCl of pH 1.0, 900 ml of distilled water is added, the pH of the resulting solution will become:

• A. 1
• B. 2
• C. 4
• D. 4.7

Answer 1

Answer: (B)

2

$pH=-log\left[H^{+}\right]=1$.
Thus $\left[H^{+}\right]={10}^{-1}$.
When 900 ml water is added to 100 ml solution, total volume becomes 1000 ml. Thus the solution is diluted to 10 times. Hence the concentration will be one tenth of original concentration.
$\left[H^{+}\right]={10}^{-2}$.
$pH=-log\left[H^{+}\right]=-log{10}^{-2}=2.0$.