Question

Calculate the pH of the resulting solution when 25.00 ml of 0.100 M H₂C₂O₄ was treated with the following volumes of 0.100 M NaOH at the following volumes ? (a) 0.00ml, (b) 15.00 ml, (c) 25.00 ml, (d) 49.9 ml ? Answers : (a) 1.26 (b) 1.86 (c) 2.88 (d) 7.39

Answer 1

H2C2O4 0.1 M and 25 ml

NaOH 0.1 M

a) NaOH volume is Zero ml

Then the oxalic acid PH is the solution PH

MV = M1V1-M2V2 (Because one is acid another one is base)

M(25+0) = (0.1x25)-(0.1x0)

25M = 2.5-0

M=0.1

PH=-log₁₀[H⁺]

PH = -log₁₀[0.1]

PH=1

b) 15 ml of NaOH

MV = M1V1-M2V2 (Because one is acid another one is base)

M(25+15) = (0.1x25)-(0.1x15)

40M = 2.5-1.5

M=0.025

PH=-log₁₀[H⁺]

PH = -log₁₀[0.025]

PH = -log₁₀[25x10^-3]

PH=3-log25

PH=1.603

C) 25 ml of NaOH

MV = M1V1-M2V2 (Because one is acid another one is base)

M(25+25) = (0.1x25)-(0.1x25)

25M = 2.5-2.5

M=0

The concentraion is Zero means it is neutral state ( PH =7)

d) 49.9 ml of NaOH

MV = M1V1-M2V2 (Because one is acid another one is base)

M(25+49.9) = (0.1x25)-(0.1x49.9)

74.9M = 2.5-4.99

74.9M=2.49 (The concentration is not having the negative sign, It indicates the basic strength is more)

M=2.49/74.9

M=0.0332 (Basic nature)

P^OH=-log₁₀[OH^-]

P^OH=-log₁₀[0.0332]

P^OH=-log₁₀[332x10^-4]

P^OH = 1.48

P^H +P^OH = 14

P^H= 14-1.48

P^H = 12.52