Two solutions, A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is
Open in App
Solution
Molarity of NaOH (4 g in 100 L)=10−3M
Molarity of H2SO4 (9.8g in 100 L)=10−3M
Equivalent of NaOH=M×V×nf=10−3×40×1=0.04
Equivalent of H2SO4=M×V×nf=10−3×10×2=0.02 MNaOH×(nf)NaOH×VNaOH−MH2SO4×VH2SO4×(nf)H2SO4=M.Vtotal 10−3×40×1−10−3×10×2=M.50 M=4×10−4 pOH=−logM =4−2log2 =3.4 pH=14−3.4=10.6