Question

Question 6 1 pts A 10.0 mL sample of 0.25 M NH3(aq) is titrated with 0.20 M HCl(aq) (adding HCl to NH3). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added acid. Ko of NH3 is 1.8 x 10-5 Henderson-Hasselbalch equation: pH = pka + log og HCI NH, NH3- Parta): 1) After adding 10 mL of the HCl solution, the mixture is (Select] the equivalence point on the titration curve. 2) The pH of the solution after adding HCl is (Select] Part b): 3) After adding 12.5 mL of the HCl solution, the mixture is (Select] the equivalence point on the titration curve. 4) The pH of the solution after adding HCl is (Select] Partc): 5) After adding 30.0 mL of the HCl solution, the mixture is (Select] the equivalence point on the titration curve. 6)) The pH of the solution after adding HCl is (Select]

Answer 1

The titration is

NH3 + HCl -------------------------------------> NH4Cl

10x0.25 0 0 intital mmoles

Kb of Nh3 = 1.8x10^-5

thus pKb of ammonia = 4.75

partA)

NH3 + HCl -------------------------------------> NH4Cl

10x0.25 0 0 intital mmoles

--- 10x0.20 --------- change

0.5 0 2 after reaction.

Thus the solution is a buffer and its pH is calculated using Hendersen equation as

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

= 4.75 + log 2/0.5= 5.352

2)and pH = 14-5.352 =8.648

1)The solution is above the equivalence point on the titration curve. { we areplotting the curve with volume of HCl on x-axis and pH on y-axis, so the starting pH is high and it decreases as titration proceeds with adition of HCl]

Part B)

NH3 + HCl -------------------------------------> NH4Cl

10x0.25 0 0 intital mmoles

--- 12.5x0.2 - change

0 0 2.5 after rxn.

3) the solution is AT THE EQUIVALENCE POINT on the titration curve.

4) pH of solution is calculated using

pH = 1/2[pKw -pKb -logC]

C = [salt] = mmoles/ total volume = 2.5/(10+12.5) =0.111M

pH = 1/2[14-4.75-log0.111] =5.102

Part C

NH3 + HCl -------------------------------------> NH4Cl

10x0.25 0 0 intital mmoles

---- 30x0.2 ---change

0 3.5 2.5

The solution has excess of strong acid and its pHis decided by [H+]

5) the mixture is WELL BELOW THE EQUIVALENCE POINT on the titration curve.

6) pH = -log[H+]

[H=] = 3.5/(10+30) =0.0875 M

pH = -log0.0875 = 1.058