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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
Calculate pH when $9.8g$ of $H_2SO_4$ is dissolved in $2L$ solution.
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Answer

$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1}{\text{Volume of solution in Litres}}$
$W_B = 9.8g$ = mass of solute
MB = Molar mass of $H_2SO_4 = 98g mol^{-1}$
Volume of solution = 2L
$\text{M}=\frac{9.8}{98}\times\frac{1}{2}=0.05\text{mol L}^{-1}$
$\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{H}^++\text{SO}^{2-}_4$
$[H^+] = 2 \times$ Molarity of $H_2SO_4$
$= 2 \times 0.05 = 0.1mol L^{-1}$
$= 10^{-1}mol L^{-1}$​​​​​​​
$\text{pH}=-\log(\text{H}^+)=-\log10^{-1}$
$=+1\log10=1\times1=1$
$[\because\log10=1]$

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