An aqueous solution is prepared by combining 50.0 mL of 0.10 M LiHC_2O_4 and 50.

Question

An aqueous solution is prepared by combining 50.0 mL of 0.10 M LiHC_2O_4 and 50.0 mL of 0.20 M NaOH? What are the equilibrium concentrations of H^+, OH', HC_2O_4^-, Li^+ and C_2O_4^2-? K_a1 = 5.4 times 10^2 and k_a2 = 5.3 times 10^-5 for H_2C_2O_4; kw = 1.0 times 10^-14 Make sure that you use acid base methodology N to solve this problem. Identify each species initially present and determine the nature of each speckles: Determine which one of the possible acid-base reaction that goes to completion can occur:

Explanation / Answer

Answer:

LiHC2O4 and NaaOH both are strong electrolytes and hence ionizes completely as,

LiHC2O4 (aq) --------> Li+ (aq) + HC2O4- (aq)

NaOH (aq) --------> Na+ (aq) + HO- (aq)

Initially,

Molarity of LiHC2O4 M1 = 0.10 M and volume V1 = 50.0 mL = 0.05 L

On mixing Molarity M2 = ? volume V2 = 50.0 + 50.0 = 100.0 mL = 0.1 L

by dilution law,

M2V2 = M1V1

M2 x .0.1 = 0.1 x 0.05

M2 = 0.05 M

i.e. [LiHC2O4] = 0.05 M

So, [Li+] = 0.05 M (As Li+ is neither acid/base this will be equilibrium concentration of Li+)

But, HC2O4-(aq) being acidic and basic, it's concentration will change depending of proportion of H+ or HO- ions.

For NaOH,

initially, M1 = 0.20 M and V1 = 0.05 L

On mixing, M2 =? and V2 = 0.1 L

by dilutuin law,

M2V2 = M1V1

M2 x 0.1 = 0.2 x 0.05

M2 = 0.1 M

i.e. [NaOH] = [HO-] = 0.1 M

HO- being a much stronger base than HC2O4- and also in double concentration hence there occurs complete neutralization.

On neutralization,

[HC2O4- ] = 0 M

[HO-]final = [HO-]initial - [HC2O4- ]initial = 0.1 - 0.05 = 0.05 M

so, [HO-] = 0.05 M

[H+][HO-] = 1 x 10-14.

[H+] x 0.05 = 1 x 10-14.

[H+] = 1 x 10-14/0.05

[H+] = 2.0 x 10-13 M.

As there is 100% neutralization of HC2O4- to C2O42- .

[C2O42-] = [HC2O4- ]initial = 0.05 M

[C2O42-] = 0.05 M.

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