Question
Calculate pH when 9.8g of H2SO4 is dissolved in 2L solution.
✓
Answer
$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1}{\text{Volume of solution in Litres}}$
WB = 9.8g = mass of solute
MB = Molar mass of H2SO4 = 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 × Molarity of H2SO4
= 2 × 0.05 = 0.1mol L-1
= 10-1mol L-1
$\text{pH}=-\log(\text{H}^+)=-\log10^{-1}$
$=+1\log10=1\times1=1$
$[\because\log10=1]$
WB = 9.8g = mass of solute
MB = Molar mass of H2SO4 = 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 × Molarity of H2SO4
= 2 × 0.05 = 0.1mol L-1
= 10-1mol L-1
$\text{pH}=-\log(\text{H}^+)=-\log10^{-1}$
$=+1\log10=1\times1=1$
$[\because\log10=1]$
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