Question

1) A solution was prepared by dissolving 489 milligrams of potassium sulfate(mw=174.24 g/mol) in 537mL of water. Calculate the following:
a) [K+]
b) % (w/v) K2SO4
c) pK+
d) pSO4^-2

Answer 1

a)

no of moles = weight in grams / molecular weight

= 0.489 / 174.24

= 0.0028 = 0.3 moles

Molarity = no of moles / volume in liters

= 0.3 / 0.537 L

= 0.5586 M

D)

K2SO4 can dissociate into

K2SO4 -----> 2K⁺ + SO₄^2-

so one mole will give 2 moles of K+

[K+] = 2 x 0.5568

= 1.1173 M

c)

pK⁺   = -log [K+]

pK+ = -log [1.1173]

pK+ = 0.0482

d)

K2SO4 can dissociate and can give one mole of SO4^2-

so concentration K2SO4 concentration of SO42-

= 0.5586 M

pSO4^-2  = -log [SO4^-2]

pSO4^-2 = -log [0.5586]

pSO4^-2 = 0.02528

b)

% (w/v) K2SO4 = weight in grams / volume in liters x 100

= 0.489 / 537

= 0.09%