25 ml sample of 0.150 m solution of aqueous trimethylamine is titrated with 0.100 M soultion of hcl. calculate the pH of the solution after 10.0ml,20 ml and 30.0 ml of acid have been added; pkb of (CH3)3 N =4.19?
To calculate the pH of the solution after different amounts of acid have been added, we need to understand the chemical reaction that is taking place between trimethylamine ((CH3)3N) and hydrochloric acid (HCl).
The balanced chemical equation for the reaction between trimethylamine and hydrochloric acid is:
(CH3)3N + HCl -> (CH3)3NH(+) + Cl(-)
From the equation, we can see that trimethylamine acts as a base and reacts with the acid, forming its conjugate acid, (CH3)3NH(+). The concentration of (CH3)3NH(+) formed will determine the pH of the solution.
To calculate the pH at different points during the titration, we need to determine the number of moles present and the final volume of the solution at each point.
Let's calculate the number of moles of trimethylamine initially present in the 25 ml of 0.150 M solution:
Number of moles of (CH3)3N = Volume (L) * Concentration (M)
= 25 ml * 0.150 M
= 0.025 L * 0.150 M
= 0.00375 moles
Now, let's calculate the remaining moles of trimethylamine after the addition of 10.0 ml, 20 ml, and 30.0 ml of the HCl solution:
Remaining moles of (CH3)3N = Initial moles - Moles of HCl added
For 10.0 ml of HCl added:
Moles of HCl = Volume (L) * Concentration (M)
= 10.0 ml * 0.100 M
= 0.010 L * 0.100 M
= 0.001 moles
Remaining moles of (CH3)3N = 0.00375 moles - 0.001 moles
= 0.00275 moles
For 20 ml of HCl added:
Moles of HCl = 20 ml * 0.100 M
= 0.020 L * 0.100 M
= 0.002 moles
Remaining moles of (CH3)3N = 0.00375 moles - 0.002 moles
= 0.00175 moles
For 30.0 ml of HCl added:
Moles of HCl = 30.0 ml * 0.100 M
= 0.030 L * 0.100 M
= 0.003 moles
Remaining moles of (CH3)3N = 0.00375 moles - 0.003 moles
= 0.00075 moles
Now that we have the remaining moles of (CH3)3N, we can calculate the concentration of (CH3)3NH(+) by dividing the moles by the final volume of the solution at each point:
Concentration of (CH3)3NH(+) = Remaining moles of (CH3)3N / Final volume (L)
Final volume = Initial volume (ml) + Volume of HCl added (ml)
For the given points:
After 10.0 ml of HCl added:
Final volume = 25 ml + 10.0 ml
Concentration of (CH3)3NH(+) = 0.00275 moles / 35 ml
= 0.00275 moles / 0.035 L
= 0.07857 M
After 20 ml of HCl added:
Final volume = 25 ml + 20 ml
Concentration of (CH3)3NH(+) = 0.00175 moles / 45 ml
= 0.00175 moles / 0.045 L
= 0.03889 M
After 30.0 ml of HCl added:
Final volume = 25 ml + 30.0 ml
Concentration of (CH3)3NH(+) = 0.00075 moles / 55 ml
= 0.00075 moles / 0.055 L
= 0.01364 M
Now, let's calculate the pH using the pKb value:
pOH = -log10(Kw / Kb)
Since pKw + pOH = 14, we can rewrite it as:
pH = 14 - pOH
For the given pKb value (4.19), we can calculate the pOH and then pH at each point:
After 10.0 ml of HCl added:
pOH = -log10(Kw / Kb)
= -log10(1.0e-14 / 4.19)
= -log10(2.392367e-16)
= 15.62
pH = 14 - pOH
= 14 - 15.62
= -1.62
After 20 ml of HCl added:
pOH = -log10(Kw / Kb)
= -log10(1.0e-14 / 4.19)
= -log10(2.392367e-16)
= 15.62
pH = 14 - pOH
= 14 - 15.62
= -1.62
After 30.0 ml of HCl added:
pOH = -log10(Kw / Kb)
= -log10(1.0e-14 / 4.19)
= -log10(2.392367e-16)
= 15.62
pH = 14 - pOH
= 14 - 15.62
= -1.62
Therefore, the pH of the solution after adding 10.0 ml, 20 ml, and 30.0 ml of HCl remains constant at -1.62.