Question

please include how you solved for the answer.

d. 7.70 7.50 e. 525.0 mL of 0.130 M NaOH is added to 525 mL of 0.250 M weak acid (Ka 6.8 x 104M). What is the pH of the resulting buffer? KG=6.8X104 PIca = 3.16 C1307 C aso 3.17 b. 3.45 a. 3.a t109 -2.91 c 3.13 d. 2.88 3.20 е.

Answer 1

Given:

M(HA) = 0.25 M

V(HA) = 525 mL

M(NaOH) = 0.13 M

V(NaOH) = 525 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.25 M * 525 mL = 131.25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.13 M * 525 mL = 68.25 mmol

We have:

mol(HA) = 131.25 mmol

mol(NaOH) = 68.25 mmol

68.25 mmol of both will react

excess HA remaining = 63 mmol

Volume of Solution = 525 + 525 = 1050 mL

[HA] = 63 mmol/1050 mL = 0.06M

[A-] = 68.25/1050 = 0.065M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.167+ log {6.5*10^-2/6*10^-2}

= 3.202

Answer: e