1 A student following the procedure in this lab prepared 4 s

1. A student following the procedure in this lab prepared 4 solutions by adding 12.25, 23.00, 25.50, and 34.50 mL of a 8.25 x 102 M NaOH solution to 50.00 mL of a 0.152 M solution of weak acid. The solutions were labeled 1, 2, 3, and 4 respectively. Each of the solutions was diluted to a total volume of 250 mL with DI water. The pH readings of these solutions were: (1) 6.64 (2) 6.98 (3) 7.04 (4) 7.23 PRE-LAB DATA SHEET 1234 Initial number of moles Number of moles at equilibrium An Equilibrium concentrations, mol/L HAnp An K, K, Which of the weak acids in Table 1 do you think your unknown acid is?

Solution

It\'s 4 buffer solutions so use H-H.

start with 7.6 mmol of acid (50 mL * 0.152 M)

Then I add base (1) 1.01 mmol (2) 1.90 mmol (3) 2.10 mmol (4) 2.85 mmol

The base neutralizes the acid, so I have the following amount of acid left:

(1) 6.6 mmol (7.6-1.01) (2) 5.7 mmol (3) 5.5 mmol (4) 4.75 mmol

The amount of conjugate base created (An-) is the same as the amount of OH- added (1.01 mmol, 1.90 mmol, etc)

the H3O+ can be calculated from the pH

pH = -log (H3O+)
so H3O+ =10^-pH
for pH = 6.64, 10^-6.64

= 2.29x10-7 M H3O+

pH = 6.98 , 1.05x10^-7 M

pH = 7.04, 9.12x10^-8 M
pH = 7.23, 5.89x10^-8

The Ka comes from Henderson-Hasselbach equation:
pH = pKa + log (base/acid)

solution 1:

6.64 = pKa + log (1.01/6.60)

pKa = 7.45

Ka=10^-7.45

= 3.51x10^-8

So the acid is HOCl