Write out the balanced chemical equations for the following titration systems: a

Question

Write out the balanced chemical equations for the following titration systems: a. Titration of HCl(aq) with NaOH(aq) b. Titration of acetic acid (CH_3COOH) with NaOH(aq) c. Titration of NH_3(aq) with HCl(aq) For the titration of 25.00 mL of 0.100 M Acetic Acid (CH_3COOH, K_a = 1.75 times 10), calculate the pH after the addition of the following volumes of 0.100 M NaOH. Show your work below and on the back of this page. 0.00 mL:____ 5.00 mL:____ 24.00 mL:____ 25.00 mL:____ 26.00 mL:____ 30.00 mL:____ 45.00 mL:____ Using a spreadsheet, create a titration curve using the above data. On the graph mark the volume and pH at the equivalence point. Attach the graph to this paper. What would be an appropriate indicator to use to measure the endpoint volume of this titration?

Explanation / Answer

Solution:-

(1)

(a) HCl(aq) + NaOH(aq) <----> NaCl(aq) + H2O(l)

(b) CH3COOH(aq) + NaOH(aq) ----> CH3COONa(aq) + H2O(l)

(c) NH3(aq) + HCl(aq) -----> NH4Cl(aq)

(2)

(i) when 0.00 ml of NaOH are added then pH will only be because of the weak acid. For weak acid or base we make the ICE table.

CH3COOH(aq) + H2O(l) <-----> CH3COO-(aq) + H3O+(aq)

I 0.100 0 0

C -X +X +X

E (0.100 - X) X X

Ka = [CH3COO-] [H3O+]/[CH3COOH]

1.75 x 10-5  = (X)2/(0.100 - X)

since the Ka is very low so the X on the bottom could be neglected.

1.75 x 10-5  = (X)2/(0.100)

on cross multiply....

X2  = 1.75 x 10-5 x 0.100 = 1.75 x 10-6

taking square root to both sides..

X = 1.32 x 10-3  = [H3O+]

pH = - log 1.32 x 10-3

pH = 2.88

(ii) when 5.00 ml of NaOH were added.

moles of NaOH added = 5.00 ml x (1L/1000ml) x (0.100mol/L) = 0.0005 mol

moles of acid = 25.00 ml x (1L/1000ml) x (0.100mol/L) = 0.0025 mol

excess moles of acid = 0.0025 - 0.0005 = 0.002 mol

moles of conjugate(salt) formed = 0.0005 mol

Total volume will be same for both the excess acid and the conjugate base so we could use the moles to calculate the pH using Handerson equation since we have a buffer solution here.

pH = Pka + log(A-/HA)

Pka = - log Ka = - log 1.75 x 10-5 = 4.76

pH = 4.76 + log(0.0005/0.002)

pH = 4.76 - 0.60

pH = 4.16

(iii) when 12.50 ml of NaOH were added..

moles of NaOH = 12.50 ml x (1L/1000ml) x (0.100mol/L) = 0.00125 mol

moles of acid = 0.0025 mol

excess moles of acid = 0.00125 mol

moles of conjugate base formed = 0.00125 mol

pH = 4.76 + log(0.00125/0.00125)

pH = 4.76

(iv) when 24.00 ml of NaOH were added..

moles of NaOH = 24.00 ml x (1L/1000ml) x (0.100mol/L) = 0.0024 mol

moles of acid = 0.0025mol

excess moles of acid = 0.0025 - 0.0024 = 0.0001 mol

moles of conjugate base formed = 0.0024 mol

pH = 4.76 + log(0.0024/0.0001)

pH = 4.76 + 1.38

pH = 6.14

(v) when 25.00 ml of NaOH were added..

25.00 ml x (1L/1000ml) x (0.100mol/L) = 0.0025 mol

moles of acid = 0.0025

here we have equal moles of both so 0.0025 moles of conjugate base are formed. Total volume is 50.00 ml = 0.0500 L.

concentration of conjugate base = 0.0025 mol/0.0500 L = 0.05 M

We will do the hydrolysis of this..

CH3COO-(aq) + H2O(l) <-----> CH3COOH(aq) + OH-(aq)

I 0.05 0 0

C -X +X +X

E (0.05 - X) X X

kb = (X)2/(0.05 - X)

kb = (1.0 x 10-14)/(1.75 x 10-5) = 5.71 x 10-10

5.71 x 10-10 = (X)2/(0.05 - X)

since Kb is very low so the X on the bottom could be neglected.

5.71 x 10-10 = (X)2/(0.05)

solving this for X..

X = 5.34 x 10-6 = [OH-]

pOH = - log 5.34 x 10-6 = 5.27

pH = 14 - 5.27

pH = 8.73

(vi) when 26.00 ml of NaOH were added..

26.00 ml x (1L/1000ml) x (0.100 mol/L) = 0.0026 mol

moles of acid = 0.0025 mol

excess moles of NaOH or OH- = 0.0026 - 0.0025 = 0.0001 mol

total volume = 51.00 ml = 0.0510 L

[OH-] = 0.0001 mol/0.0510 L = 0.00196 M

pOH = - log 0.00196 = 2.71

pH = 14 - 2.71

pH = 11.29

(vii) when 30.00 ml of NaOH were added..

30.00 ml x (1L/1000ml) x (0.100mol/L) = 0.0030 mol

moles of acid = 0.0025 mol

excess moles of NaOH or OH- = 0.0030 - 0.0025 = 0.0005 mol

total volume = 25.00 ml + 30.00 ml = 55.00 ml = 0.055 L

[OH-] = 0.0005 mol/0.055 L = 0.00909 M

pOH = - log 0.00909 = 2.04

pH = 14 - 2.04

pH = 11.96

(viii) when 45.00 ml of NaOH were added..

45.00 ml x (1L/1000ml) x (0.100mol/L) = 0.0045 mol

moles of acid = 0.0025 mol

excess moles of NaOH or OH- = 0.0045 - 0.0025 = 0.0020 mol

total volume = 25.00ml + 45.00ml = 70.00 ml = 0.070 L

[OH-] = 0.0020 mol/0.070 L = 0.0286 M

pOH = - log 0.0286 = 1.54

pH = 14 - 1.54

pH = 12.46

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