Question

A 200ml sample of the lake water is collected titrated directly with 0.01 M NaOH and it takes only 3ml of NaOH titrant to each the equivalent point. calculate the molarity (M) of HNO3 in the lake water.

Answer 1

c) the reaction is given by

HN03 + NaOH ---> NaN03 + H20

now

we can see that

at equivalence point

moles of NaOH added = moles of HN03 present

now

we know that

moles = molarity x volume

so

molarity x volume of NaOH added = molarity x volume of HN03

so

Mb xVb = Ma x Va

given

molarity of NaOH ( Mb) = 0.01

Volume of NaOH ( Vb) = 3

volume of HN03 (Va) = 200

so

Mb xVb = Ma x Va

0.01 x 3 = Ma x 200

Ma = 1.5 x 10-4

so

the molarity of HN03 in lake water is 1.5 x 10-4 M

d)

given

conc = 0.08 mg / ml

= 0.08 x 10-3 g / 10-4 L

conc = 0.08 g / L

now

conc in mol / L = conc in g / L / molar mass

aslo

molar mass of Hn03 = 63 g/mol

so

conc of Hn03 in mol/L = 0.08 / 63

conc of HN03 in mol/L = 1.27 x 10-3

now

HN03 + H20 --> H30+ + N03-

we can see that

[H30+] = [HN03]

so

[H30+] = 1.27 x 10-3

now

we know that

pH = -log [H30+]

so

pH = -log 1.27 x 10-3

pH = 2.90

so

the pH of lakewater is 2.90