Question

a) How many moles of water are formed when 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320 M KOH are mixed?

b) What is the concentration of H+(aq) and OH–(aq) after the reaction is complete?

Answer 1

a)

for HNO3:

volume , V = 28.0 mL

= 0.028 L

number of mol,

n = Molarity * Volume

= 0.25*0.028

= 0.007 mol

volume , V = 53.0 mL

= 0.053 L

FOR KOH:

number of mol,

n = Molarity * Volume

= 0.32*0.053

= 0.017 mol

Balanced chemical equation is:

HNO3 + KOH ---> H2O + KNO3

1 mol of HNO3 reacts with 1 mol of KOH

for 0.007 mol of HNO3, 0.007 mol of KOH is required

But we have 0.017 mol of KOH

so, HNO3 is limiting reagent

we will use HNO3 in further calculation

According to balanced equation

mol of H2O formed = moles of HNO3

= 0.007

= 0.007 mol

Answer: 0.007 mol

b)

According to balanced equation

mol of KOH formed = (1/1)* moles of HNO3

= (1/1)*0.007

= 0.007 mol

mol of KOH remaining = mol initially present - mol reacted

mol of KOH remaining = 0.017 - 0.007

mol of KOH remaining = 0.01 mol

total volume = 0.028 L + 0.053 L = 0.081 L

[OH-] = 0.01 mol / 0.081 L

= 0.123 M

[H+] = 1.0*10^-14 / [OH-]

[H+] = 1.0*10^-14 / 0.123

[H+] = 8.13*10^-14

Answer:

[H+] = 8.13*10^-14

[OH-] = 0.123 M