Question

A mixture is prepared using 27.00 mL of a 0.0758 M weak base (pKa=4.594) , 27.00 mL of a 0.0553 M weak acid (pKa=3.235) and 1.00 mL of 1.25×10−4 M HIn and then diluting to 100.0 mL, where HIn is the protonated indicator. The absorbance measured at 550 nm in a 5.000 cm cell was 0.1102 . The molar absorptivity (????) values for HIn and its deprotonated form In− at 550 nm are 2.26×104 M−1cm−1 and 1.53×104 M−1cm−1 , respectively. What is the pH of the solution?

Answer 1

Answer:

pH = 4.164

Explanation:

The first process is to find the initial moles for the base (B) & the acid (HA)

i.e.

$= \dfrac{27 mL \times 0.0758 \ moles \ \ of \ B}{1000 \ mL}$

$=0.0020466$

$\simeq 2.047\times 10^{-3} \ moles \ of \ B$

$= \dfrac{27 mL \times 0.0553 \ moles \ \ of \ HA}{1000 \ mL}$

$=0.0014931$

$\simeq 1.493\times 10^{-3} \ moles \ of \ HA$

The acid with base reaction is expressed as;

          HA              +      B         →             A⁻            +         HB⁺

to    1.493 × 10⁻³       2.047 × 10⁻³             -                         -

    -  1.493 × 10⁻³       1.493 × 10⁻³         1.493 × 10⁻³         1.493 × 10⁻³

            0                    5.54 × 10⁻⁴         1.493 × 10⁻³         1.493 × 10⁻³

From observation; both the acid & base weak

Given that:

The pKa for base = 4.594

The pKa for acid = 3.235

Recall that;

pKa = -log Ka

So; Ka = $\mathbf{10^{-Ka}}$

By applying this:

For Base; Ka = $10^{-4.594}$ = 2.5468 × 10⁻⁵

For Acid:  Ka = $10^{ -3.235}$ = 5.821 × 10⁻⁴

After the reaction; we have the base with its conjugate acid & conjugate base of acid; Thus, since the conjugate acid of the base possesses a higher value of K, it is likely it would be the one to define the pH of the solution.

By analyzing the system, we have:

              HB⁺         + H₂O           ↔           B          +            H₃O⁺

     $\dfrac{1.493\times 10^{-3}}{0.1 \ L}$                                 $\dfrac{5.54\times 10^{-4}}{0.1 \ L}$

to   0.01493 M                                 0.00554   M

  -      x                                                   x                                x

    0.01493 - x                                 0.00554 - x                     x

Thus;

$2.5468 \times 10^{-5} = \dfrac{(0.00554 -x)\times x}{(0.01493-x)}$

Using the common ion effect;

0.00554 - x   $\simeq$  0.00554   &

0.01493 -  x   $\simeq$   0.01493

∴

$x = \dfrac{2.5468 \times 10^{-5} \times 0.01493}{0.00554}$

x =  [H₃O⁺] = 6.8635  × 10⁻⁵

∴

pH = -log(6.8635  × 10⁻⁵)

pH = 4.164