Question

please help with number 4. if you feel like hwlping with number 5, that would be cool too. thank you

E) 1812.a 4) A 15 mL sample of 0.025 M NH3 is titrated with a 0.016 M solution of HCl. What is the pH after 12 mL of acid is added? The Kb of NH3 is 1.8.10-5 B) 10.27 C) 2.09 4 O 9210 5) A 17 mL sample of 0.25 M glacial acetic acid is titrated with a 0.25M solution of NaOH. What is the pH half way to the equivalence point? Ka-1.8 10-5? A) 3.45 B) 6.09 C) 4.74 D) 5.28

Answer 1

Given:

M(HCl) = 0.016 M

V(HCl) = 12 mL

M(NH3) = 0.025 M

V(NH3) = 15 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.016 M * 12 mL = 0.192 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.025 M * 15 mL = 0.375 mmol

We have:

mol(HCl) = 0.192 mmol

mol(NH3) = 0.375 mmol

0.192 mmol of both will react

excess NH3 remaining = 0.183 mmol

Volume of Solution = 12 + 15 = 27 mL

[NH3] = 0.183 mmol/27 mL = 0.0068 M

[NH4+] = 0.192 mmol/27 mL = 0.0071 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {7.111*10^-3/6.778*10^-3}

= 4.76

use:

PH = 14 - pOH

= 14 - 4.76

= 9.24

Answer: D