Explanation

HN$O_3$ + KOH -> KN$O_3$ + $H_2$ O

Mole of NHO$_3$ = 0.2moldm$^-3$ x ($50/1000$ )dm$^3$

=0.01mol

mole of KOH=0.32moldm$^-3$ x (25/1000) dm$^3$

=0.008mol

since 1 m0le of KOH will require 1mole of HNO$_3$ , mole of HNO$_3$ required to react with 0.008 mol of KOH=0.08mol.

Thus, mole of unreacted HNO$_3$

=(0.01-0.008)mol = 0.002mol.

Total volume = 50cm$^3$ + 25cm$^3$

=75cm$^3$

=(75/1000)dm$^3$

=0.075dm$^3$

thus concentration of unreacted HNO$_3$

=0.002mol/0.0075dm$^3$

=0.0267mold$m^{-3}$

HNO$_3$ -> H$^+$ + NO$^-$ $_3$

0.0267 0.0267

pH= -log$_10$ (H$^+$ )

=-log(0.0267)

=1.574

HN$O_3$ + KOH -> KN$O_3$ + $H_2$ O

Mole of NHO$_3$ = 0.2moldm$^-3$ x ($50/1000$ )dm$^3$

=0.01mol

mole of KOH=0.32moldm$^-3$ x (25/1000) dm$^3$

=0.008mol

since 1 m0le of KOH will require 1mole of HNO$_3$ , mole of HNO$_3$ required to react with 0.008 mol of KOH=0.08mol.

Thus, mole of unreacted HNO$_3$

=(0.01-0.008)mol = 0.002mol.

Total volume = 50cm$^3$ + 25cm$^3$

=75cm$^3$

=(75/1000)dm$^3$

=0.075dm$^3$

thus concentration of unreacted HNO$_3$

=0.002mol/0.0075dm$^3$

=0.0267mold$m^{-3}$

HNO$_3$ -> H$^+$ + NO$^-$ $_3$

0.0267 0.0267

pH= -log$_10$ (H$^+$ )

=-log(0.0267)

=1.574