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  4. When 80 mL of 0.20 M HCl is mixed with 120 mL of 0.15 M KOH, the resultant solution is the same as a solution of

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Q. When $80\, mL$ of $0.20\, M\, HCl$ is mixed with $120 \,mL$ of $0.15 \,M\, KOH$, the resultant solution is the same as a solution of

Some Basic Concepts of Chemistry

Solution:

$80\, mL$ of $0.20 \, M \, HCl =80 \times 0.2$
$=16$ millimol $HCl$
$120 \, mL$ of $0.15 \, M\, KOH =120 \times 0.15$
$= 18$ millimol $KOH$
$M_1 V_1( HCl ) < M_2 V_2( KOH )$
Thus, resulting solution is basic containing $KCl$ and unreacted $KOH$
$KOH + HCl \rightarrow KCl + H _2 O$
$KOH =\frac{M_2 V_2-M_1 V_1}{V_1+V_2}=\frac{18-16}{200}=0.01 \, M$
$KCl$ formed $= HCl$ used
$=16$ millimoles
$=16 \times 10^{-3} mol$ in $200 \, mL$ or $0.2 \, L$ solution
$[ KCl ]=\frac{16 \times 10^{-3}}{0.200\, L }=0.08\, M$