A solution is prepared by mixing together, 35.0 mL 0.200 M HCl 10.0 mL 0.150 M H

Question

A solution is prepared by mixing together,

            35.0 mL 0.200 M HCl

            10.0 mL 0.150 M HClO4

            25.0 mL 0.300 M NaOH

            20.0 mL 0.250 M NH3

            10.0 mL water

Assume volumes are additive.

(1) Identify each compound as an acid or base, and write appropriate dissociation reactions, including the proper arrow notation. Determine the millimoles of each solute initially present and whether the overall solution will be acidic or basic.

(2) Determine the pH of the solution at 25 °C.

(3) Determine how much 0.400 M HClO4 or 0.200 M NaOH must be added to give a final solution pH of 9.30. Include an explanation of your reasoning on this part of the problem.

Explanation / Answer

Solution:-

(1) These acids and bases dissociates as follows...

HCl(aq) ---------> H+(aq) + Cl-(aq)

HClO4(aq) --------> H+(aq) + ClO4-(aq)

NaOH(aq) ----------> Na+(aq) + OH-(aq)

NH3(aq) + H2O(l) <-----------> NH4+(aq) + OH-(aq)    ammonia is a weak base so It's good to write its equation with water to show how it produces OH- in the solution.

from the given initial concentrations and volumes we can calculate the moles.

initial mmol of HCl = 0.200 x 35.0 = 7.0

mmol of HClO4 = 0.150 x 10 = 1.5

mmol of NaOH = 0.300 x 25 = 7.5

mmol of NaOH = 0.250 x 20 = 5.0

so, total mmol of H+ = 7.0 + 1.5 = 8.5

mmol of OH- = 7.5

pH calculations in part (2) shows that the over all solution is basic.

(2) these H+ and OH- react then excess mmol of H+ = 8.5 - 7.5 = 1.0

mmol of NH3 are 5.0 and it would react with 1.0 mmol of H+ to give 1.0 mmol of NH4+ and 4.0 mmol of NH3 be in excess and so we have a buffer solution. Final volume would be same for NH3 as well as NH4+ so we could use the mmol in Handerson equation to calculate the pH.

Pkb for NH3 is 4.74. we would use the Handerson equation to calculate the pH.

Pka = 14 - Pka = 14 - 4.74 = 9.26

pH = Pka + log(base/acid)

pH = 9.26 + log(4.0/1.0)

pH = 9.26 + 0.60

pH = 9.86

So, the over all solution is basic.

(3) Initially the pH was 9.86 and now the new pH is 9.30

Here the pH is decreasing from 9.86 to 9.30, it inidicates that an acid is added, since the addition of acid decreases the pH where as the addition of base increases it. So, let's say X mmol of acid are added to the buffer.

These X mmol of acid would react with X mmol of NH3 to form X mmol of NH4+

so, new mmol of NH3 = 4 - X

new mmol of NH4+ = 1 + X

Let's again plug in the values in Handerson equation and calculate the value of X

9.30 = 9.26 + log[(4-X)/(1+X)]

subtract 9.26 from both sides..

0.04 = log[(4-X)/(1+X)]

taking antilog..

100.04 = [(4-X)/(1+X)]

1.096 = [(4-X)/(1+X)]

on cross multiply...

1.096(1+X) = (4-X)

1.096 + 1.096X = 4 - X

on arrangement...

1.096X + X = 4 - 1.096

2.096X = 2.904

X = 2.904/2.096 = 1.385

It means we need to add 1.385 mmol of HClO4.

Molarity of HClO4 is given as 0.400M

we know that, mmol = molarity x volume in ml

so, volume in ml = mmol/molarity = 1.385/0.400 = 3.4625 ml

So, we need to approxymately 3.46 ml of 0.400 M HClO4

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