Question

3] You are charged with producing a buffer with a pH 7.65, by dissolving a weighed amount of sodium hypochlorite (NaCIO, FM-74.44) in 1.5 L of 0.35 M hypchlorous acid (HOCI, pka 7.46) (a) What mass (in g) of NaCIO is required (4 points (b) What is the resulting pH if 20 mL of 0.08 M HCI is added to 80 mL of this buffer? (4 points (c) Could i make a second NaCIO/HOCl buffer with a pH = 5.0? Explain. (2 points) Problem IV: Titrations 1] In order to titrate 24.5 mL of a HNO3 solution, 33.15 mL of 0.1053 M NaOH is needed. What is the pH of this solution when only 20.22 mL of the NaOH titrant had been added· points

Answer 1

Answer 3 (a) - Mass of NaOCl:

Henderson–Hasselbalch equation for Weak acid/conjugate base buffer equilibrium:

pH = pK_a + log ([A^-]/[HA])

where [A^-] = molar concentration of a conjugate base

[HA] = molar concentration of a undissociated weak acid (M)

for Buffer (HOCl/NaOCl) of pH= 7.65

pH = pKa + log [NaOCl]/[HOCl]

7.65 = 7.46 + log [NaOCl]/[HOCl]

0.19 = log [NaOCl]/[HOCl]

Taking Antilog

1.548817 = [NaOCl]/[HOCl]

So [NaOCl] = 1.55 Times of [HOCl]= 1.55 * 0.35 M = 0.542 M

Since the mol wt of NaClO = 74.44

Quantity of NaClO required for preparing 1.5 Lt of buffer = 1.5* 74.44 * 0.542 = 60.51 gm

Answer 3 (b) : Resulting pH when 20 ml of 0.08 M HCl is added to 80 ml of this buffer:

We know know the molarity of [NaOCl] = 0.542 M and [HOCl] = 0.35 M

80 ml of the buffer NaOCl/HOCl will contain mols of

mols of NaOCl = (80/1000)* 0.542 = 43.36 * 10^-3 = 0.0433mols

mols of HOCl = (80/1000)* 0.35 = 28 * 10^-3 = 0.028mols

When we are adding 20 ml of 0.08 M HCl = .0016 mols of H+

HOCl + NaOH $\rightleftharpoons$ NaOCl + H₃O⁺

It will react with NaOCl to increse the [HOCl] by 0.0016 mols, by doing this [NaOCl] will decrease by 0.0016 mols

Now,.ols of NaOCl= 0.0433mols - 0.0016 mols = 0.0417 mols

mols of HOCl= 0.028mols + 0.0016 mols=0.0296 mols

The New PH as per Henderson–Hasselbalch equation

pH = pKa + log [NaOCl]/[HOCl]

    = 7.46 + log [NaOCl]/[HOCl]

     = 7.46 + log (0.0417)/(0.0296)

   = 7.46 + log (1.4) = 7.46 + 0.146

     pH = 7.606 (after adding 20 ml of 0.08 M HCl).

Note: total volume = 80+20 = 100 ml will be same for both [NaOCl] and [HOCl] which will be cancelled out when putting the values of [ ] in moles/Lt.

(c) For preparing the second ph 5 buffer, we need to add the excess amount of HCL in to the buffer.

pH = pKa + log [NaOCl]/[HOCl]

5 = 7.46 + log [NaOCl]/[HOCl]

log [NaOCl]/[HOCl] = -2.46

or log [HOCl]/[NaOCl] = 2.46

taking antilog [HOCl]/[NaOCl] = 1.30

So, for a HOCl/NaOCl buffer to have pH 5, we need to add a lot of HCL so that the concentration of HOCL becomes 1.3 times the conce

[HOCl] = 1.3 * [NaOCl]

if we know the initial concentration of [HOCl] and [NaOCl] and [HCL], We can calculate the amount (Volume) of HCL needed to male the [HOCl] = 1.3 * [NaOCl] for preparing the buffer of pH 5.

Answer No. IV :

First we calculate the Concentration of HNO₃ solution.

24.5 (C1) = 33.15 (0.1053)

C1 = 0.1424 M

So initially we have 24.5 ml of 0.1424 M HNO₃.

mols of HNO₃ = ₍24.5 ml/ 1000)* 0.1424 = 0.0035 mols

For, 20.22 ml 0.1053 M of NaOH

mols of NaOH= ₍20.22 ml/ 1000)* 0.1053 = 0.00212 mols

So moles of HNO3 left, after adding 20.22 ml 0.1053 M of NaOH = 0.0035 mols- 0.00212 mols = 0.00138 mols

Total volume = 20.22 ml+ 24.5 ml = 64.72 ml

$\therefore$ Molarity of HNO3 (after adding 20.22 ml 0.1053 M of NaOH) = 0.00138 /(64.72/1000) = 0.021 M

pH of 0.021 M HNO3

pH = -log [H⁺]

     = - log [0.021] = -(-1.677)

pH = 1.677