Question

10 mL of a strong acid solution of pH = 2.000 is mixed with 990 mL of another strong acid solution of pH = 4.000. What will be the pH of the resulting solution?
log1.99=0.3

• A. 4.002
• B. 4000
• C. 4.2
• D. 3.7

Answer 1

The correct option is D 3.7

$\left[H^{+}\right]$ after mixing $\frac{{10}^{-2}\times 10+{10}^{-4}\times 990}{1000}=\frac{0.1+0.0990}{1000}=\frac{0.1990}{1000}=1.99\times {10}^{-4}pH=-(\log 1.99\times {10}^{-4})$
$pH=(4-\log 1.99)$
$pH=4-0.3=3.7$