Question

$400mL$ of a strong acid solution of pH = 3 is mixed with $600mL$ of another strong base solution of pOH = 4. The pH of resulting solution will be :
[Given, $\log \left(3.4\right)=0.531$]

• A. 3.08
• B. 4
• C. 4.2
• D. 3.46

Answer 1

The correct option is D 3.46


$p{H}_{solution1}=3;C_1=\left[H^{+}\right]={10}^{-3}pO{H}_{solution2}=4;C_2=\left[O{H}^{-}\right]={10}^{-4}{C}_f=[H^{+}{]}_{\text{remains after neutralization}}{V}_1=400mL,V_2=600mL{V}_f=V_1+V_2=1000mLApplying:$
$C_1{V}_1-C_2{V}_2=C_f{V}_f{C}_f=\frac{C_1{V}_1-C_2{V}_2}{V_f}=\frac{({10}^{-3}\times 400)-({10}^{-4}\times 600)}{1000}=3.4\times {10}^{-4}pH=-{\log }_{10}[3.4\times {10}^{-4}]pH=3.46$