2
in question 1 part d rearrange the expression for kwkwkw to work out the ohtextohoh ion concentration

a calculate the ph of a solution containing 0100 mol dm30100 textmol dm30100mol dm3 ethanoic acid and
0100 mol dm30100 textmol dm30100mol dm3

Question

2.
In Question 1 part d, rearrange the expression for $K_{w}$ Kw to work out the ${OH}^{-}$ OH− ion concentration.

a) Calculate the pH of a solution containing $0.100 {mol dm}^{- 3}$ 0.100mol dm−3 ethanoic acid and $0.100 {mol dm}^{- 3}$ 0.100mol dm−3 sodium ethanoate?
( $K_{a}$ Ka of ${CH}_{3} COOH = 1.74 \times 1 0^{- 5} {mol dm}^{- 3}$ CH3COOH=1.74×10−5mol dm−3) [3]

b) How many moles of sodium ethanoate must be added to $2.00 {dm}^{3}$ 2.00dm3 of $0.0100 {mol dm}^{- 3}$ 0.0100mol dm−3 ethanoic acid to produce a buffer solution of pH 5.40? [5]

2 in question 1 part d rearrange the expression for k to work out the oh ion concentration a calculate the ph of a solution containing 0100 mol dm2 ethanoic acid and 0100 mol dm3 sodium ethanoate k of ch3cooh 174 x 107 mol dm3 31 b how many moles of sodium ethanoate must be added to 200 dm of 00100 mol dm3 ethanoic acid to produce a buffer solution of ph 540 5

Abel Varughese · Answer

a) Calculate the pH of a solution containing 0.100 mol dm⁻³ ethanoic acid and 0.100 mol dm⁻³ sodium ethanoate

We use the Henderson-Hasselbalch equation for buffer solutions:

$pH = pKa + \log (\frac{[Salt]}{[Acid]})$

Step 1: Calculate pKa
The given $K_{a}$ of ethanoic acid is $1.74 \times 1 0^{- 5} {mol dm}^{- 3}$ .

$pKa = - \log (K_{a}) = - \log (1.74 \times 1 0^{- 5}) \approx 4.76$

Step 2: Substitute values
The concentrations of the acid and salt are equal ( $[Acid] = [Salt] = 0.100 {mol dm}^{- 3}$ ).

$pH = 4.76 + \log (\frac{0.100}{0.100})$ $pH = 4.76 + \log (1)$ $pH = 4.76$

Answer for part a:
The pH of the solution is 4.76.

b) How many moles of sodium ethanoate must be added to 2.00 dm³ of 0.0100 mol dm⁻³ ethanoic acid to produce a buffer solution of pH 5.40?

Step 1: Use the Henderson-Hasselbalch equation
Rearrange the equation to solve for the salt concentration:

$pH = pKa + \log (\frac{[Salt]}{[Acid]})$ $\log (\frac{[Salt]}{[Acid]}) = pH - pKa$ $\frac{[Salt]}{[Acid]} = 1 0^{(pH - pKa)}$

Substitute the values $pH = 5.40$ , $pKa = 4.76$ , and $[Acid] = 0.0100 {mol dm}^{- 3}$ :

$\frac{[Salt]}{0.0100} = 1 0^{(5.40 - 4.76)} = 1 0^{0.64} \approx 4.37$ $[Salt] = 4.37 \times 0.0100 = 0.0437 {mol dm}^{- 3}$

Step 2: Calculate moles of sodium ethanoate required
The total volume of the solution is $2.00 {dm}^{3}$ . The moles of sodium ethanoate are:

$Moles of Salt = [Salt] \times Volume = 0.0437 {mol dm}^{- 3} \times 2.00 {dm}^{3}$ $Moles of Salt = 0.0874 mol$

Answer for part b:
The amount of sodium ethanoate required is 0.0874 moles.

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