100mL of 0.1M NaOH solution is titrated with 100mL of 0.05MH2SO4 solution. The pH of the resulting solution is : (forH2SO4,Ka1=∞,Ka2=10−2)
A
7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
7.2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
7.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6.8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B 7.2 no of milimoles of NaOH=100×0.1=10
no of milimoles of H2SO4=100×0.05=5
now 2NaOH+H2SO4→Na2SO4+2H2O
Since 1 mole H2SO4 rects with 2 mole of NaOH
the above reaction will go till complete neutralization and after that [SO4]2−=5200M and this will go anion hydrolysis SO2−4+H2O⇋HSO−4+OH−
Where HSO−4 is conjugate acid with Ka=10−2 pH=12[pKw+pKa+logC] pH=12[14+2−1.6]=7.2