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consider a strong acid - strong base titration OF 35.00 mL sample of 0.175 m HBr...

consider a strong acid - strong base titration OF 35.00 mL sample of 0.175 m HBr with 0.200 m KOH

what is the ph of a solution after 10.00 ml of the KOH has been added? Stichiometry only, no equilibrium

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Answer #1

no. of mole = molarity X volume of solution in liter

no. of moel of KOH = 0.200 X 0.010 = 0.0020 mole

no. of mole of HBr = 0.175 X 0.035 = 0.006125 mole

neutrilization reaction is

HBr + KOH KBr + H2O

According to neutrilization reaction between HBr and KOH they react equimolary therefore to react with 0.0020 mole of KOH required HBr = 0.0020 mole

mole of HBr remain in solution after complete reaction = 0.006125 - 0.0020 = 0.004125 mole

total volume of solution = 35 + 10 = 45 ml = 0.045 L

molarity = no. of mole/volume of solution in liter

molarity of HBr after adding 10 ml KOH = 0.004125 / 0.045 = 0.09167 M

HBr is strong aid dissociate completly therefore [HBr] = [H3O+] = 0.09167 M

pH = -log[H3O+]

pH of solution = -log(0.09167) = 1.037

pH of soluition = 1.037

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