Question

0.100 molar weak acid HA vs 0.100 molar NaOH Start with 25.0 mL HA I will provide a blank graph if you want to use it to fill in the points. You can use a pencil and take a picture) or 'draw'. Or you can use a graphing program. Please answer questions, 2 points each According to this data: What is the pka of the acid? What is [H*) at the Equivalence point? What is the [OH-] at the Equivalence point? What is the [HA) at the equivalence point? What is the [H+) at the Half Equivalency point? I Page 2 of 2 ma 174 words Focus 6:34 PM 7/31/2020 12 ODO D Home End Insert OOFTO! o F9 F11 F12

Answer 1

1. pKa of the weak acid is equal to pH at half equivalence point. From the graph and table, the equivalence point is 25 mL. So, the half equivalence point is 25/2=12.5 mL. Now, we have to check the pH at this point which is 4.24.

Hence, pKa = 4.24

2. As the pH at equivalence point (25 mL) is equal to 8.46 i.e. pH = 8.46

pH = -log[H⁺]

[H⁺] = 10^-pH = 10^-8.46 = 3.47x10^-9 M

Hence, at equivalence point, [H⁺] = 3.47x10^-9 M

3. As we know that

[H⁺][OH^-] = K_w

So, at eqivalence point

OH ] H+1

$[OH^-] = \frac{1.0\times 10^{-14}}{3.47\times 10^{-9}} = 2.88\times 10^{-6} M$

Hence, at equivalence point, [OH^-] = 2.88x10^-6 M

4. At equivalence point, hydrolysis of conjugate base of weak acid (A^-) takes place by the below reaction,

A^-(aq) + H₂O(l) $\rightleftharpoons$ HA(aq) + OH^-(aq)

So, the concentration of HA is equal to concentration of [OH^-] at equivalence point.

[HA] = [OH^-] = 2.88x10^-6 M

Hence, at equivalence point, [HA] = 2.88x10^-6 M

5. As the pH at half equivalence point is 4.24. So,

pH = -log[H⁺]

[H⁺] = 10^-pH = 10^-4.24 = 5.75x10^-5 M

Hence, at half equivalence point, [H⁺] = 5.75x10^-5 M

Let me know if you have any queries regarding this.