TA 1.22 g sample of pure monoprotic acid, HA, was dissolved in di a mL of soluti

Question

TA 1.22 g sample of pure monoprotic acid, HA, was dissolved in di a mL of solution. The HA solution was then titrated with 0.250 MNaOH. The pH was measured throughout the titration, and the equivalence point had been added. The data from the itration are recorded in the table belowt stilled water to make 50.0 was reached when 40.0 mL of the NaOH solution added. The data from the titration are recorded in the table below. olume of 0.250 MNaOHpH of Titrated Added (mL) 0.00 10.0 20.0 30.0 40.0 50.0 Solution 3.72 4.20 8.62 12.40 () (3) Explain how the data in the table above provide evidence that HA is a weak acid rather than a strong acid. (b) [3] Write the balanced net-ionic equation for the reaction that occurs when the solution of NaOH is added to the solution of HA . -2 (e) (6] Calculate the number of moles of HA that were titrated and the molar mass of HA. (d) [11] The equilibrium constant for HA is K-6.3 × 10-5 M. Calculate the initial concentration the HA solution (before any NaOH solution was added) and determine the pH of the initial HA solution. (Be sure to write an equation for the dissociation reaction of HA and set up the ICE table

Explanation / Answer

a.

At end point (40 mL) pH is basic (8.62) corresponding to a solution that contains the weak base A- conjugated to the weak acid AH.

If AH is a strong acid, at end point pH=7 (this is not the case here).

b.

HA + HO- = A- + H2O

c.

The number of moles of HA = the number of moles of NaOH used in this titration

               0.250 mol/L x 0.040 L = 0.010 mol

The molar mass of HA

                M = m/ 0.010 mol

                     = 1.22 g/ 0.010 = 122 g/mol

d.

The initial concentration of HA was

             0.010 mol/0.050 L = 0.2 mol/L

HA    =     A-   +   H+

0.2 M……0……..0             initial

-x……….+x……+x            change

0.2-x…...+x……+x           at equilibrium

Ka = x2/(0.2-x)

Assume that x << 0.2 and solve for x.

x = 3.55x10-3 M

[H+] = x = 3.55x10-3M

pH = -log[H+] = 2.45

e.

At 20 mL , the half end point where

pH = pKa + log([A-]/[AH]) and [A-]=[AH]      

f. At 30 mL

pH = pKa + log([A-]/[AH]) and [A-]/[AH] = 0.75/025 = 3

pH = pKa + log3

      = 4.20 +0.48 = 4.68

g. Choose thymol blue (at end point pH=8.62). It will change the colour from yellow to blue.

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