Question

Given this information:

Solution #8 is 50mL 0.10M NH3 + 50mL 0.10M NH4NO3

Solution #10= 10mL Solution 8 + 5mL H2O + 1mL 0.10M HCl

I’m being asked use the ICF calculations (with moles, not molarity) and use the Henderson-Hasselbach equation to find the pH for solution #10

I’m so lost and this is due in less than three hours. Please help

Answer 1

Solution 8 : 50 ml 0.1 M ammonia + 50 ml 0.1 M ammonium nitrate

pH of above solution is given by using Henderson's equation as,

pH = p K _a + log [ NH₃ ] / [ NH₄⁺]

No of moles of NH₃ = Molarity x volume of solution in litre

= 0.1 x 0.05

= 5 x 10 ^-3 mol

No of moles of NH₄⁺= Molarity x volume of solution in litre

   = 0.1 x 0.05

= 5 x 10 ^-3 mol

Therefore, pH = 9.2 + log 5 x 10 ^-3 / 5 x 10 ^-3

= 9.2 + log 1

= 9.2 + 0

= 9.2

Here volume of solution is 100 ml .

Therefore, 100 ml buffer solution contain 5 x 10 ^-3 mol of ammonia.

Hence,10 ml buffer solution contain ( 5 x 10 ^-3 x 10 / 100 ) = 5 x 10 ^-4 mol ammonia.

Similarly, moles of NH₄⁺ in 10 ml buffer solution = 5 x 10 ^-4 mol

No of moles of HCl = Molarity x volume of solution in ltre

=0.1 x 0.001

= 1 x 10 ^-4 mol

HCl is a acid . When it is added to Buffer solution it combines with basic component of buffer as shown below.

NH₃ + H ⁺ NH₄⁺

i e concentration of base decreases and that of acid increases.

Therefore , no of moles of NH₃ = 5 x 10 ^-4 - 1 x 10 ^-4 = 4 x 10 ^-4 mol

no of moles of NH₄⁺ =  5 x 10 ^-4 +1 x 10 ^-4= 6 x 10 ^-4 mol

Therefore, pH of buffer after adding HCl is

pH = 9.2 + log 4 x 10 ^-4 / 6 x 10 ^-4

= 9.2 -0.18

= 9.02