A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H5O2. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HC7H5O2 is 6.5 × 10-5. A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H5O2. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HC7H5O2 is 6.5 × 10-5. 3.34 3.97 5.03 4.41 4.19

By adding the strong acid HCl to the buffer, some of the conjugate base of the weak acid will be consumed and some of the weak acid will produced through the chemical equation:

HCl + LiC7H5O2 ⇒ HC7H5O2 + LiCl

Therefore we need to account for this reaction to answer the question:

Vol HCl = 100 mL = 0.100 L

mol HCl reacted = 0.100 L x 1.00 mol/L = 0.100 mol

mol LiC7H5O2 originally present = 0.250 mol/L x 1.00 L = 0.250 mol

mol LiC7H5O2 after reaction with HCl = 0.250 -0.100 M = 0.250 mol

mol HC7H5O2 originally present = 0.150 mol/L x 1.00 L = 0.150 mol

mol HC7H5O2 produced in reaction = 0.100 mol

mol HC7H5O2 present after reaction = 0.150 mol + 0.100 mol = 0.250 mol

Now we are in position to calculate the pH by plugging our values into the equation:

pH = pKa + log ((A⁻) / (HA)) = 4.20 + log (0.15/0.25) = 4.20 + log 0.60

= 3.97

Note: We do need to calculate concentration since the volume is 1 L and do not affect the ratio (A⁻) / (HA) and cancel out.

AFOKE88 AFOKE88 · Answer 2 · 2022-03-18T19:00:42+01:00

The PH of the Solution is; PH = 3.96

What is the PH of the Solution?

Let us call the benzoic acid C₆H₅COOH to be HA

It's manner of dissociation is;

HA ⇄ H⁺ (aq) + A⁻ (aq) ------(eq 1)

The lithium salt LiC₇H₅O₂ provides a large reserve of the co - base

A⁻ by dissociating completely:

LiA (s) → Li⁺ (aq) + A⁻ (aq) -----(eq 2)

The initial number of moles of A⁻ is given by:

nA⁻_init = 1 * 0.25 = 0.25

The number of moles of Hydrogen ions added is;

nH⁺ = 1 * (100/1000) = 0.1

Since H⁺ and A⁻ ions react in a 1:1 molar ratio, then;

Number of moles of A⁻ left is; nA⁻ = 0.25 - 0.1 = 0.15

While number of moles of H⁺ formed = 0.1 moles

Total moles of HA is;

nHA = 0.1 + 0.15

nHA = 0.25

For the formula for the acid dissociation constant, we can get [H⁺ (aq)] as;

[H⁺ (aq)] = K_a * [HA (aq)]/[A⁻ (aq)]

Plugging in the relevant values;

[H⁺ (aq)] = 6.5 * 10⁵ * (0.25/0.15)

[H⁺ (aq)] = 1.083 * 10⁻⁴

pH = -log[1.083 * 10⁻⁴]

pH = 3.96

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