Question

A 2.461 g sample of solid sodium acetate is dissolved in enough water to make 50.00 mL of solution. To the sodium acetate solution, 100 mL of 0.120 M HCl is added. What is the pH of the resulting solution. $(K_a for CH_{3}COOH$ = 1.8 \times 10^{-5}[/tex])

Answer 1

"From the given,

Given amount of sodium acetate = 2.461 g

Molar mass of sodium acetate = 82.0 g/mol

Number of moles of sodium acetate added $=\quad \frac { 2.461\quad g }{ 82.03\quad g/mol } \quad =\quad 3.00\quad \times \quad { 10 }^{ -2 }\quad moles\quad =\quad 0.0300\quad mole$

$C{ H }_{ 3 }CO{ O }^{ - }(aq)\quad +\quad { H }_{ 3 }{ O }^{ + }(aq)\quad \leftrightarrow \quad C{ H }_{ 3 }COOH(aq)\quad +\quad { H }_{ 2 }O$

Number of moles of ${ H }_{ 3 }{ O }^{ + }$ added = (0.1000 L) (0.120 M) = 0.0120 mole

From the above equation,

0.0120 mol ${ H }_{ 3 }{ O }^{ + }$ react with 0.0120 mole of ${ CH }_{ 3 }CO{ O }^{ - }$ to form 0.0120 mole $C{ H }_{ 3 }COOH$

The number of moles of acetate ion in excess is = 0.0300 - 0.012 = 0.0180 mole

The solution after mixing the two reagents contains 0.0120 mole of ${ CH }_{ 3 }COOH$ and 0.0180 mole of ${ CH }_{ 3 }{ COO }^{ - }$ in a total volume of 0.1500 L.

$[{ CH }_{ 3 }COOH]\quad =\quad \frac { 0.0120\quad mol }{ 0.1500\quad L } \quad =\quad 0.0800\quad M$

$[{ CH }_{ 3 }CO{ O }^{ - }]\quad =\quad \frac { 0.0180\quad mol }{ 0.1500\quad L } \quad =\quad 0.1200\quad M$

${ K }_{ a }\quad ({ CH }_{ 3 }COOH)\quad =\quad 1.8\quad \times \quad { 10 }^{ -5 }$

$p{ K }_{ a }\quad =\quad -log1.8\quad \times \quad { 10 }^{ -5 }$

$=\quad 5\quad -\quad log1.8$

$p{ K }_{ a }\quad =\quad 4.74$

$pH\quad =\quad p{ K }_{ a }\quad +\quad log\frac { [{ CH }_{ 3 }CO{ O }^{ - }] }{ [{ CH }_{ 3 }COOH] }$

$=\quad 4.74\quad +\quad log\left( \frac { 1.20 }{ 0.0800 }  \right)$

$pH\quad =\quad 5.92$"

Answer 2

Heya mate....☺☺

$\huge{\bold{\green{pH=5.92}}}$

I HOPE ITS HELPFUL...☺