Question

In: Chemistry

A 25.6 mL sample of 0.219 M trimethylamine, (CH3)3N, is titrated with 0.373 M hydrochloric acid....

A 25.6 mL sample of 0.219 M trimethylamine, (CH₃)₃N, is titrated with 0.373 M hydrochloric acid.

After adding 5.62 mL of hydrochloric acid, the pH is .

Expert Solution

The reaction between (CH3)3N and HCl is

(CH3)3N + HCl - - - - - - > (CH3)3NH+ + Cl^-

1:1 molar reaction

Initial moles of (CH3)3N = (0.219mol/1000ml)× 25.6ml = 0.0056064

No of moles of HCl added = (0.373mol/1000ml)×5.62ml = 0.002096 mol

0.002096moles of HCl react with 0.002096moles of (CH3)3N to produce 0.002096moles of (CH3)3NH+

After adding HCl

Volume of solution = 25.6ml + 5.62ml = 31.22ml

[(CH3)3NH+] = (0.002096mol/31.22ml)×1000ml = 0.0671M

[(CH3)3N]= ((0.0056064mol- 0.002096mol)/31.22ml)×1000ml = 0.1124M

pKa of (CH3)3NH+ = 9.81

Henderson-Hasselbalch equation is

pH = pKa + log([A^-] /[HA])

pH = 9.81 + log([(CH3)3N)]/[(CH3)3NH+])

pH = 9.81 + log(0.1124M/0.0671M)

pH = 9.81 + 0.22

pH = 10.03

queen_honey_blossom answered 2 years ago

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