Abeaker with 1.80×102 ml of an acetic acid buffer with a ph of 5.000 is sitting on a benchtop. the total molarity of acid and conjugate base in this buffer is 0.100 m. a student adds 6.50 ml of a 0.330 m hcl solution to the beaker. how much will the ph change? the pka of acetic acid is 4.740.

The pH change in 0,206 units

Explanation:

When the acetic acid buffer is at pH 5,000; it is possible to obtain the acetate/acetic acid proportions using Henderson-Hasselbalch formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] Where A⁻ is CH₃COO⁻ and HA is CH₃COOH.

Replacing:

5,000 = 4,740 + log₁₀ [A⁻]/[HA]

1,820 = [A⁻]/[HA] (1)

As buffer concentration is 0,100M:

[A⁻] + [HA] = 0,100 (2)

Replacing (2) in (1)

[HA] = 0,035M

And [A⁻] = 0,065M

As volume is 1,80x10²mL, moles of HA and A⁻ are:

0,180L × 0,035M = 6,3x10⁻³mol of HA

0,180L × 0,065M = 1,17x10⁻²mol of A⁻

The reaction of HCl with A⁻ is:

HCl + A⁻ → HA + Cl⁻

The add moles of HCl are:

0,0065L×0,330M = 2,145x10⁻³ moles of HCl that are equivalent to moles of A⁻ consumed and moles of HA produced.

Thus, moles of HA after addition of HCl are:

6,3x10⁻³mol + 2,145x10⁻³ mol = 8,445x10⁻³ moles of HA

And moles of A⁻ are:

1,17x10⁻²mol - 2,145x10⁻³ mol = 9,555x10⁻³ moles of A⁻

Replacing these values in Henderson-Hasselbalch formula:

pH = 4,740 + log₁₀ [9,555x10⁻³ ]/[8,445x10⁻³ ]

pH = 4,794

The pH change in 5,000-4,794 = 0,206 units

I hope it helps!

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

hope i !