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
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