Question

The hydrogen ion concentration and pH of the solution obtained by mixing 100 mi of 1.0M HNO3 with 100 ml of 0.8M KOH is:

Answer 1

Given,
100mL of 1.0M of HNO3
we know, mole = concentration × volume
= 100 × 1.0 = 100milimole

similarly, 100mL of 0.8M of KOH,
mole of KOH = 100 × 0.8 = 80 millimole

see the reaction of HNO3 and KOH,
HNO3 + KOH = KNO3 + H2O
here we can see that 1 mole of KOH is neutralised by 1 mole of HNO3.

so, 80 millimole of KOH is neutralised by 80 millimole of HNO3,
hence, rest mole of HNO3 in reaction = 100 - 80 = 20 millimole.

now, concentration of rest HNO3 = mole of rest HNO3/volume
= 20/(100 + 100) = 20/200 = 0.1M
hence, concentration of H^+ = 0.1M

pH = -log[H^+] = -log(10^-1) = 1

Answer 2

Given,

100mL of 1.0M of HNO3

we know, mole = concentration × volume

= 100 × 1.0 = 100milimole

similarly, 100mL of 0.8M of KOH,

mole of KOH = 100 × 0.8 = 80 millimole

see the reaction of HNO3 and KOH,

HNO3 + KOH = KNO3 + H2O

here we can see that 1 mole of KOH is neutralised by 1 mole of HNO3.

so, 80 millimole of KOH is neutralised by 80 millimole of HNO3,

hence, rest mole of HNO3 in reaction = 100 - 80 = 20 millimole.

now, concentration of rest HNO3 = mole of rest HNO3/volume

= 20/(100 + 100) = 20/200 = 0.1M

hence, concentration of H^+ = 0.1M

pH = -log[H^+] = -log(10^-1) = 1

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