2. A buffer solution contains 0.120 M acetic acid and 0.150 M sodium a cetat a.

Question

2. A buffer solution contains 0.120 M acetic acid and 0.150 M sodium a cetat a. H L of ow many moles of acetic acid and of sodium acetate are present in 50.0 m solution? how many moles of resent after the reaction is done? b. If we add 5.55 mL of 0.092 M NaOH to the solution in part a., acetic acid, sodium acetate, and NaOH will be p If we add 0.50 mL of 0.087 M HCI to the solution in part a., how many moles of acetic acid, sodium acetate, and HCI will be present after the reaction is done? c. 3. Determine the phH you expect to find for the three solutions in Question 2 ine the pH you expect to find for the three solutions in Question 2. 3. Determi

Explanation / Answer

2) a) moles of CH3COOH = molarity x volume in lItres

= 0.12 M x 0.05 L

= 0.006 mol

moles of CH3COONa = molarity x volume in lItres

= 0.15 M x 0.05 L

= 0.0075 mol

b) moles of NaOH =  molarity x volume in lItres

= 0.092 M x 0.00555 L = 0.00051 mol

CH3COOH + NaOH ---------------------> CH3COONa + H2O

0.006 mol 0.00051 mol 0

-----------------------------------------------------------------------------------------------

0.006 - 0.00051 0 0.00051 mol

= 0.00549 mol

Hence, after the reaction done

[CH3COOH]= 0.00549 mol

[CH3COONa] = 0.0075 mol + 0.00051 mol = 0.00801 mol

[NaOH] = 0 mol

c)

moles of HCl =  molarity x volume in lItres

= 0.087 M x 0.0005 L = 0.0000435 mol

CH3COONa + HCl ---------------------> CH3COOH + NaCl

0.0075 mol 0.0000435 mol 0

----------------------------------------------------------------------------------------------------------

0.0075 -0.0000435 0 0.0000435 mol

= 0.0074565 mol

Hence, after the reaction done

[CH3COOH]= 0.006 mol + 0.0000435 mol = 0.0060435 mol

[CH3COONa] = 0.0074565 mol

[HCl] = 0 mol

3)

pKa of acetic acid = 4.8

pH = pKa + log [CH3COONa]/ [CH3COOH]

a) pH = 4.8 + log (0.0075)/(0.006) = 4.9

b) pH = 4.8 + log (0.00801 mol)/(0.00549 mol) = 4.96

c) pH = 4.8 + log (0.0074565 mol)/( 0.0060435 mol) = 4.89

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