Answer :
Volume required is 41.625 ml of 0.20M NaOH must be added to the 20 ml of 5%. acetic acid solution to bring the pH to 4.74.
Solution -
Step 1: To calculate molarity of acetic acid in100 mL of solution
Molar mass of acetic acid = 60.0591 mol
No. of moles of acetic acid = mass of acetic acid ÷Molar mass of acetic acid
=5g÷60.05g/mol
moles of acetic acid 0.08326 mol
Molarity of acetic acid =moles of acetic acid/ Vol of sol" (in L)
=0.08326 mol/L÷0.1L
=0.8326mol/L
Molarity of 20 mL of 5% acetic acid solution = (20 mL) x 0.8326 mol /L
= 0.01665 mol/L
Step 2 :To calculate volume of 0.20 M NaOH required.
Let 'v' liters of 0.20 M NaOH was added to the acetic acid solution.
New Molarity of NaOH, x = (V x 0·20) M -equation (1)
Given: pH=4.74 ,pka=4.74
Henderson-Hasselbalch equation is:
[tex]pH = pK_a + log10([A^-] / [HA])[/tex]
[A⁻] = conjugate base
[HA] = acid
Acetic acid + NaOH → Sodium acetate + H2O
Initial 0.01665 M xM 0 M
change -x -x +x
Final 0.01665-x 0 xM
pH=pK a +log [Sodium acetate] [acetic acid]
4 * 74 = 4 * 74 + log(x/(0.01665 - x))
0 = log(x/(0.01665 - x))
Taking antilog both
[antilog (0) = 1]
1 = x / 0.016C5-x
0.01665 - x = x
x = 0.008325 M
New So, Molarity of NaOH = 0.008325 M
From equation (1).
New Molarity of NaOH = V ×0.20M
=0.008325 V × 0.20
V= 0.041625 L
V= 41.625 mL
Thus the volume will be V= 41.625 mL
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