Question

$500mL$ of $0.1molHCl$ has $pH$

• A. 1
• B. 13
• C. 2
• D. 0.1

Answer 1

Answer: (A)

$\left[H^{+}\right]+0.1/500/1000$
$=\frac{0.1\times 1000}{500}=0.2$
$pH=-\log \left[H^{+}\right]$
$=-[\log 0.2]=-\log \left[2\times 10^{-1}\right]$
$=-[\log 2-1\log 10]$
$=-\log 2+1$
$=-0.3010+1=0.6990$
$\approx 1.$