Answer:
pOH = 5.961
Explanation:
To find the pH of a weak base we can use Henderson-Hasselbalch equation for weak bases:
pOH = pKb + log [(CH₃)₃NHCl] / [(CH₃)₃N]
Where pKb is -log Kb = 4.187 and [] could be taken as moles of each specie.
Moles (CH₃)₃NHCl:
0.0441L * (0.15mol/L) = 6.615x10⁻³moles
Moles (CH₃)₃N:
0.0233L * (0.16mol/L) = 3.728x10⁻³moles
And pOH is:
pOH = pKb + log [(CH₃)₃NHCl] / [(CH₃)₃N]
pOH = 4.187 + log [6.615x10⁻³moles] / [3.728x10⁻³moles]
The pOH of the solution is 4.44
Data;
The pOH of a solution is calculated by
[tex]pOH = -logk_b + log \frac{[(CH3)3N]}{[CH3NH3Cl]}[/tex]
Let's find the actual concentration of the solutions
m mole of (CH3)3N = 23.3 * 0.16 = 3.728 mmol
m mol of CH3NH3Cl = 44.1 * 0.15 = 6.615 mmol
The pOH of the solution becomes
[tex]pOH = -logk_b + log \frac{[(CH3)3N]}{[CH3NH3Cl]}\\pOH = -log(6.5*10^-^5) + log (\frac{6.615}{3.728} \\pOH = -log (6.5*10^-^5) + 0.249\\pOH = 4.44[/tex]
The pOH of the solution is 4.44
Learn on pH and pOH of a solution here;
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