Question

What mass of a weak acid with a molar mass of 100 g/mol is necessary to neutralize 25 ml of 0.10 M NaOH solution?

What is the pH of 0.15 g of sodium acetate NaC2H3O2 in 100 ml water H2O?

Answer 1

a)

moles of NaOH = molarity * vol = 0.10*25/1000 = 2.5*10^-3

                                    = moles ofweak acid = mass/mol mass

   So           2.5*10^-3 = mass/100

so mass = 2.5*10^-3 * 100   = 0.25 g

b)    Kb for sodium acetate is 5.65*10^-10

molar conc = 0.15(mass) /mol mass /vol in litres    = 0.15/82*100/1000 = 0.0183

Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:

Ac¯ + H₂O <==> HAc + OH¯

1             1             1       1

0.0183-x                  x      x

2) Here is the K_b expression for Ac¯:

	[HAc] [OH¯]
Kb =	----------------
	[Ac¯]

3) We can then substitute values into the K_b expression in the normal manner:

	(x) (x)
5.65 x 10¯10 =	----------------
	0.0.0183 - x

4) Ignoring the minus x in the usual manner, we proceed to solve for the hydroxide ion concentration:

x = square root of [(5.65 x 10¯¹⁰) *0.0183

x =6.78*10^-6 M = [OH¯]

We then calculate the pH. Since this is a base calculation, we need to do the pOH first:

pOH = - log 6.78*10^-6 = 5.16

pH = 14 - 5..16 = 8.8310