Question

Consider a strong acid HA having a pH=4 mixed in an equal volume with another solution of strong acid HB to give a resultant pH of 3. Then the $H^{+}$ concentration of solution containing strong acid HB is :

• A. $3.5\times {10}^{-3}M$
• B. $4.9\times {10}^{-2}M$
• C. $1.9\times {10}^{-3}M$
• D. $2.2\times {10}^{-5}M$

Answer 1

Answer: (C)

$1.9\times {10}^{-3}M$

Solution 1 having strong acid HA:
$pH=4-log\left[H^{+}\right]=4\left[H^{+}\right]={10}^{-4}{C}_1={10}^{-4}M$
Solution B having strong acid HB:
Let concentration be $C_2$
Final solution obtained after mixing
$pH=3-log\left[H^{+}\right]=3\left[H^{+}\right]={10}^{-3}M$

Concentration $\left(C_f\right)={10}^{-3}M$
The volume of two solutions are equal.
So,
$V_1=V_2=V{V}_f=V_1+V_2=2V{C}_f{V}_f=C_1{V}_1+C_2{V}_2{10}^{-3}\times 2V=({10}^{-4}\times V)+(C_2\times V)C_2=(2\times {10}^{-3})-\left({10}^{-4}\right)C_2=1.9\times {10}^{-3}M$