Question

1. A 30.00 mL sample of 0.400 M HNO3 is titrated with 0.600 M KOH. What is the balanced neutralization chemical reaction? What volume of base (in mL) must be added to reach the equivalence point? Is the pH of the equivalence point acidic, basic, or neutral?

Answer 1

the balanced equation is as follows

HNO3 + KOH -----------------> KNO3 + H2O

no of moles = molarity * volume in L

here forHNO3 => 0.030 L * 0.400 M => 0.012 moles

the mole ratio of the HNO3 & NaOH are 1:1

using the dilution law

M1V1 = M2V2

for HNO3 => M1 = 0.400 M , V1 = 0.030 L

for KOH => M2 = 0.600 M , V2 = ?

V2 = M1V1 / M2

V2 = 0.400 * 0.030 / 0.600

V2 = 0.020 L

V2 = 20.0 mL

C) HNO3 strong acid and H+ => 0.400 M

pH = -log[H+]

pH = 0.39794 at equivalence point is acidic