Ionic Equilibrium Question 7
Question 7 - 30 January - Shift 1
$600 mL$ of $0.01 M HCl$ is mixed with $400 mL$ of $0.01 M H_2 SO_4$. The $pH$ of the mixture is $\times 10^{-2}$. (Nearest integer)
[Given $\log 2=0.30, \quad \log 3=0.48$
$\log 5=0.69$
$\log 7=0.84$
$\log 11=1.04]$
Show Answer
Answer: (186)
Solution:
Formula: pH claculation of strong acid
Total milimoles of $H^{+}=(600 \times 0.01)+(400 \times 0.01 \times 2)$
$ \begin{gathered} =14 \\ {[H^{+}]=\frac{14}{1000}=14 \times 10^{-3}} \\ pH=3-\log 14 \\ =1.86 \\ =186 \times 10^{-2} \end{gathered} $
