Question

Part A volume of 40.0 mL of a 0.820 M HNO3 solution is titrated with 0.390 M KOH. Calculate the volume of KOH required to reach the equivalence point Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) Value Units Submit

Answer 1

HNO₃ + KOH -----> KNO₃ + H₂O

Number of moles of HNO3 = molarity * volume of solution in L

Number of moles of HNO3 = 0.820 * 0.04 = 0.0328 mole

From the balanced equation we can say that

1 mole of HNO3 requires 1 mole of KOH so

0.0328 mole of HNO3 will require

= 0.0328 mole of HNO3 *(1 mole of KOH / 1 mole of HNO3)

= 0.0328 mole of KOH

molarity of KOH = number of moles of KOH / volume of solution in L

0.390 = 0.0328 / volume of solution in L

volume of solution in L = 0.0328 / 0.390 = 0.0841 L

1 L = 1000 mL

0.0841 L = 84.1 mL

Therefore, the volume of KOH required would be 84.1 mL