Question

A 25.0 ml sample of 0.150 m butanoic acid is titrated with a 0.150 m naoh solution. what is the ph after 26.0 ml of base is added? the ka of butanoic acid is 1.5 ⋅ 10-5. a 25.0 ml sample of 0.150 m butanoic acid is titrated with a 0.150 m naoh solution. what is the ph after 26.0 ml of base is added? the ka of butanoic acid is 1.5 10-5. 4.84 2.54 7.00 4.81 11.47

Answer 1

The pH of the solution after adding 26.0 mL of 0.150 M NaOH to a 25.0 mL sample of 0.150 M butanoic acid is 4.81.

Step-by-step explanation:

The pH of the solution after adding 26.0 mL of 0.150 M NaOH to a 25.0 mL sample of 0.150 M butanoic acid can be calculated using the Henderson-Hasselbalch equation. This equation is used to calculate the pH of a weak acid solution. The Henderson-Hasselbalch equation is given by:

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

In this equation, pKa is the negative logarithm of the acid dissociation constant (Ka). For butanoic acid, the Ka value is 1.5 x 10^-5. [A-] is the concentration of the conjugate base (butanoate ion) and [HA] is the concentration of the acid (butanoic acid).

Using the given values, the concentration of the butanoate ion [A-] can be calculated as follows:

[A-] = [NaOH] x V(NaOH) / V(total)

where [NaOH] is the concentration of the NaOH solution, V(NaOH) is the volume of NaOH added (26.0 mL), and V(total) is the total volume of the solution (25.0 mL).

Substituting the values into the equation, [A-] = (0.150 M) x (26.0 mL) / (25.0 mL + 26.0 mL), we get [A-] = 0.079 M.

The concentration of the acid [HA] can be calculated as follows:

[HA] = initial concentration - [A-] = 0.150 M - 0.079 M = 0.071 M.

Substituting the values into the Henderson-Hasselbalch equation, we get:

pH = -log(1.5 x 10^-5) + log (0.079 M / 0.071 M) = 4.81.

Therefore, the pH of the solution after adding 26.0 mL of 0.150 M NaOH is 4.81.

Answer 2

first,
Moles of HBu = Volume per liter * moles of HBu/ liter
= 25ml/1000 * 0.15 = 0.00375 moles
moles of NaOH = volume per liter * moles of NaOH/liter
= 1 ml/1000 * 0.15 = 0.00015 moles
according to this equation:
HBu + NaOH → H2O + NaBu
when 1 mol of NaOH gives 1mol of HBu
So 0.00015 of NaOH will give 0.00015 mol of HBu
∴moles of HBu which remains = 0.00375- 0.00015 = 0.0036 moles

∴moles of Bu- produced = 0.00015 moles
when the total volume = 0.025 + 0.026 =0.051 L
[HBu] = 0.0036moles / 0.051 L = 0.071 moles
[Bu] = 0.00015 / 0.051L = 0.0029 moles
when Ka = [H+] [Bu] / [HBu]
1.5x10^-5 = [H+] (0.0029) /(0.071)
∴[H+] =1x10^-6 / 0.076 = 1.5 x 10^-5
∴PH = -㏒[H+]
= - ㏒ 1.5x10^-5
= 4.81