Question

10 ml of 0.2 M NaOH is added to 50 ml of 0.2M of the hypothetical acid HA. write the chemical equation and calculate the molarity of NaH and HA in the final solution . Keq=1.76*10^-5

Answer 1

V = 10 ml

M = 0.2 M of NaOH

V = 50 ml

M = 0.2 HA

a)

equation

HA(aq) + NaOH(aq) ---> H2O(l) + NaA(aq)

b)

molarity of NaH and HA

NaH = mol of NaH/V

(choose the least, since there is limiting reactant)

mol of Na = M*V = 10*0.2 = 2 mmol

Then

[NAH] = 2 mmol / (50+10 ml) = 0.03333 M

for HA

[HA] = mmol of HA / VT

VT = 50 + 10 = 60 ml

mmol of HA = initial - reacted = 50*0.2-10*0.2 = 8 mmol left

then

[HA] = 8/60 = 0.13333 M

The pH

this is a buffer so

pH = pKa +log(NaH/HA)

pH = -log(1.76*10^-5) + log(0.03333/0.13333 ) = 4.15239