40ml of 0.01M NaOH are added to 10ml of 0.45M HCL. What is the PH of the resulting solution.[Kw=1*10^-14]

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NaOH + HCl ------> NaCl + H2O

no. of moles of NaOH = molarity X volume in litres = 0.01 X 0.04 = 0.0004

no. of moles of HCl = 0.45 X 0.01 = 0.0045

since HCl amount is greater than that of NaOH so NaOH is the limiting reagent ...in other words since according to reaction HCl and NaOH are reacting in 1:1 mole ratio ....so some amount of HCl will be left because it is present in greater quantity ...


amount of HCl left = 0.0045 - 0.0004 = 0.0041

total volume of solution = 40 + 10 = 50 ml = 0.05 L

molarity of HCl left = 0.0041/0.05 = 0.082 M

since HCl -----> H+ + Cl-

so [HCl] = [H+] = 0.082

pH = -log [H+] = -log 0.082 = 1.086

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