Question

Hey guys, I am having a lot of trouble with my homework. Could you please help? I thumbs up anyone who gives a thorough explanation and shows the work! I really need to understand these concepts!

1) At 25 °C, the Ka for formic acid (HCO2H) is 1.8 x 10-4. What is the pH of a 0.10 M aqueous solution of lithium formate (LiHCO2)? 2) What is the pH (aq., 25 °C) of this solution: adding 1.64 grams of sodium acetate to give 200.0 ml solution at 25.0°C? The Ka at 25.0°C for acetic acid is 1.8 x 10-5.

Thanks!

Answer 1

Answer 1:

LiHCO2 will be reacted with water by following reacton

LiHCO2 + H2O -----> HCO2H + OH^- + LI⁺

Now this is the dissociation of the OH- ion from water molecules.

so we need to use the Kb of th reaction

Suppose x moles of LiHCO2 dissociate the x mole of OH- than

Kb = [OH-] [HCOOH] / [LiHCO2]

now

Kb =10^(-14)/Ka = 10^(-14) /1.8*10^(-4) = 5.56*10^(-11)

[LiHCO2] = 0.1 - x

[OH-] =x

[HCOOH] = x

5.56*10^(-11) = x*x/(0.1-x)

We can remove the value of x with respect to 0.1

x^2 = 0.1*5.56*10^(-11)

x = 2.36*10^(-0.6)

pOH =-log[OH-] = -log(2.36*10^(-6)) = 5.6276

pH = 14-pOH = 14-5.6276 =8.3723

So pH of 0.1 M of LiHCO2 is 8.3723

Answer 2:

Moleculare weight of CH3COONa is 82.03 g/ mole

now if we add 1.64 g so moles of sodium acetate = 1.64 /82.03 = 0.02 moles

0.02 moles of sodium acetate in 200 ml so in 1000 ml it should be 0.02*5 = 0.1 moles

so molarity of the solution is 0.1 M

NaHCO2 will be reacted with water by following reacton

NaHCO2 + H2O -----> HCO2H + OH^- + LI⁺

Now this is the dissociation of the OH- ion from water molecules.

so we need to use the Kb of th reaction

Suppose x moles of NaHCO2 dissociate the x mole of OH- than

Kb = [OH-] [HCOOH] / [NaHCO2]

now

Kb =10^(-14)/Ka = 10^(-14) /1.8*10^(-5) = 5.56*10^(-10)

[NaHCO2] = 0.1 - x

[OH-] =x

[HCOOH] = x

5.56*10^(-10) = x*x/(0.1-x)

We can remove the value of x with respect to 0.1

x^2 = 0.1*5.56*10^(-10)

x = 7.45*10^(-0.6)

pOH =-log[OH-] = -log(7.45*10^(-6)) = 5.1276

pH = 14-pOH = 14-5.1276 =8.8723

So pH of 0.1 M of NaHCO2 is 8.8723

So pH of 1.64 g of sodium acetate in 200 ml will have the pH of the NaHCO2 is 8.8723