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Question

Two solutions, A and B each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

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Solution

Molarity of NaOH (4 g in 100 L)=103 M
Molarity of H2SO4 (9.8g in 100 L)=103 M
Equivalent of NaOH=M×V×nf=103×40×1=0.04
Equivalent of H2SO4=M×V×nf=103×10×2=0.02
MNaOH×(nf)NaOH×VNaOHMH2SO4×VH2SO4×(nf)H2SO4=M.Vtotal
103×40×1103×10×2=M.50
M=4×104
pOH=log M
=42log2
=3.4
pH=143.4=10.6

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