A 42.5-mL sample of 0.205 M CH_3 COOH solution is titrated with 0.172 M NaOH. Ca
ID: 1013500 • Letter: A
Question
A 42.5-mL sample of 0.205 M CH_3 COOH solution is titrated with 0.172 M NaOH. Calculate the pH of the solution (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) a before the addition of any NaOH solution pH = b after the addition of 14.1 mL of NaOH solution pH = c after the addition of 22.2 mL of NaOH solution pH = d after the addition of 24.5 mL of NaOH solution pH = e after the addition of 28.1 mL of NaOH solution pH = f after the addition of 50.7 mL of NaOH solution pH = g after the addition of 51.8 mL of NaOH solution pH =Explanation / Answer
a) before addition of NaOH
pH of weak acid = 1/2(pka-logC))
= 1/2(4.74-log0.205)
= 2.714
b) afteraddition of 14.1 mlof NaOH
No of mol of NaOH = 14.1/1000*0.172 = 0.00242 mol
No of mol of CH3COOH = 42.5/1000*0.205 = 0.0087 mol
pH = pka + log(salt/acid)
= 4.74 + log(0.00242/(0.0087-0.00242))
= 4.32
c)
afteraddition of 22.2 mlof NaOH
No of mol of NaOH = 22.2/1000*0.172 = 0.00382 mol
No of mol of CH3COOH = 42.5/1000*0.205 = 0.0087 mol
pH = pka + log(salt/acid)
= 4.74 + log(0.00382/(0.0087-0.00382))
= 4.63
d)
afteraddition of 24.5 mlof NaOH
No of mol of NaOH = 24.5/1000*0.172 = 0.004214 mol
No of mol of CH3COOH = 42.5/1000*0.205 = 0.0087 mol
pH = pka + log(salt/acid)
= 4.74 + log(0.004214/(0.0087-0.004214))
= 4.713
e)
afteraddition of 28.1 mlof NaOH
No of mol of NaOH = 28.1/1000*0.172 = 0.004833 mol
No of mol of CH3COOH = 42.5/1000*0.205 = 0.0087 mol
pH = pka + log(salt/acid)
= 4.74 + log(0.004833/(0.0087-0.004833))
= 4.84
f)
afteraddition of 50.7 mlof NaOH
No of mol of NaOH = 50.7/1000*0.172 = 0.00872 mol
No of mol of CH3COOH = 42.5/1000*0.205 = 0.0087 mol
concentration of salt = (0.0087/(50.7+42.5))*1000 = 0.093 M
pH =7+1/2(pka+logC)
= 7+1/2(4.74+log0.093)
= 8.85
g) concentration of excess NaOH = ((51.8 - 50.7)/102.5)*0.172
= 0.0018 M
pH = 14-(-log(OH-))
= 14-(-log0.0018)
=11.25
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