Question

1) a) Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate. (b) Calculate the pH of a buffer formed by mixing 85 mL of 0.13 M lactic acid with 95 mL of 0.15M sodium lactate.

2) You are asked to prepare a pH=3.00 buffer solution starting from 1.25 L of a 1.00 M solution of hydroflouric acid (HF) and any amount you need of sodium flouride (NaF). (a) What is the pH of the hydroflouric acid solution prior to adding sodium flouride? (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

3) Calculate the solubility of Mn (OH)2 in grams per liter when buffered at pH (a) 7.0, (b) 9.5, (c) 11.8

Answer 1

the Ka value of lactic acid () = 1.4 * 10-4

we use the Henderson Hasselbach equation:
pH = pKa + log( [salt] / [Acid] )

pKa = -log (ka) = -log (1.4x10^-4) = 3.8538
[acid] = concentration of lactic acid = .12M
[salt = concentration of salt = .11M
putting value in abnove equation

pH = 3.8538 + log [.11 / .12] = 3.8538 + log (0.9166) = 3.8538 -0.0378 = 3.816

2) What is the pH of the hydroflouric acid solution prior to adding sodium flouride t means we have to deal with HF only

   HF <------> H+ + F-   

initial 1 0 0

equilibrium 1-x x x

as Ka = [H+]*[F-]/[HF]
Set [H+] = [F-] = [x]
Ka = [x]*[x] / [1.0 - x]
Since x is assumed to be very small, 1.0 - x is approximately equal to 1.0. as HF is weak electrolyte
Ka = [x] * [x] / 1.0 = [x]²

[x] = (Ka) = (6.8*10^-4) = 2.61*10^-2 M
[H+] = 2.61*10^-2 M
pH = -log[H+] = -log[2.61*10^-2] or   pH = 1.58

the pH found to be without any addition of NaF

(b) as   HF <-------> H+ + F-
using above equation
pKa = pH - log([F-]/[HF]))
log([F-]/[HF])) = pH - pKa
pKa = -log(6.8*10^-4) = 3.17
as we know that pH = 3.00
log([F-]/[HF])) = 3.00 - 3.17 = -0.17

We know that [HF] = 1.00 M
log([F-]/[1])) = -0.17
log[F-] = -0.17
[F-] = 10^-(0.17)
c= 0.676 M concentration of [F-]
Molarity = no of moles of solute / dissolve per litre of the soltion
n = M * V = 0.676 M * 1.25 L = 0,845 mol F-
mass NaF = 0.845 mol * 42 g/mol
mass NaF = 35.5 g

( why you have taken sodium benzoate ?) if there is sodium benzoate then mutiply by the molar mass of

sodium benzoate to  0.845 mol to get the mass of  sodium benzoate .

3)as the equatio is given by Mn(OH)2 <------------> Mn+2 + 2OH-

Ksp=[Mn+2][OH-]² =
solubility (x) = [Mn+2]=

Ksp/[OH-]^2 =Ksp/ (10^-pOH)^2 ( in this step we have multiply and divide OH by log to convert it into pOH)

= Ksp/10^-2pOH= Ksp /(10^-2*(14-pH))

a) x =(1.6*10^-13)/(10^-(2*(14-7))) = 16 M

b) x =(1.6*10^-13)/(10^-(2*(14-9.5))) = 1.6*10^-4 M

c) x =(1.6*10^-13)/(10^-(2*(14-11.8))) =4.0* 10^-9 M