A solution is made by combining 300mL of 0.020M trimethyl amine, that is (CH3)3N

Question

A solution is made by combining 300mL of 0.020M trimethyl amine, that is (CH3)3N and a 500mL of a 0.03M solution of it's conjugate base, that is (CH3)3NH+. The Kb for trimethyl amine is 6.3x10-5 Now, what is the ratio of conjugate base to acid (that is what is the A term in the Henderson-Hasselbalch equation?) HA Question 4 1 pts A solution is made by combining 300mL of 0.020M trimethyl amine, that is (CH3)3N and a 500mL of a 0.03M solution of it's conjugate base, that is (CH3)3NH+. The Kb for trimethyl amine is 6.3x10-5 What is the pH of the solution?

Explanation / Answer

3)

Concentration after mixing = mol of component / (total volume)

M((CH3)3N) after mixing = M((CH3)3N)*V((CH3)3N)/(total volume)

M((CH3)3N) after mixing = 0.02 M*300.0 mL/(300.0+500.0)mL

M((CH3)3N) after mixing = 7.5*10^-3 M

Concentration after mixing = mol of component / (total volume)

M((CH3)3NH+) after mixing = M((CH3)3NH+)*V((CH3)3NH+)/(total volume)

M((CH3)3NH+) after mixing = 0.03 M*500.0 mL/(500.0+300.0)mL

M((CH3)3NH+) after mixing = 1.875*10^-2 M

[A]/[HA] =(7.5*10^-3)/(1.875*10^-2)

= 0.40

Answer: 0.40

4)

Kb = 6.3*10^-5

pKb = - log (Kb)

= - log(6.3*10^-5)

= 4.201

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

pOH = pKb + log {[HA]/[A]}

= 4.201+ log {1/0.40}

= 4.6

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.6

= 9.4

Answer: 9.4

Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know

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