Calculate the change in pH that occurs when 1.00 mmol of a strong acid is added

Question

Calculate the change in pH that occurs when 1.00 mmol of a strong acid is added to 100 mL of the solutions listed below. (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.) H_2O Change in pH = 0.0510 M HCL Change in pH = 0.0510 M NaOH Change in pH = 0.0510 M CH_3COOH Change in pH = 0.0510 M CH_3COONa. Change in pH = 0.0510 M CH_3COOH + 0.0510MCH_3COONa Change in pH = 0.510 M CH_3COOH + 0.510 M CH_3COONa. Change in pH =

Explanation / Answer

Moles of strong acid = 1mmol=1*10-3 moles

when acid is added , concentration of acid in 100ml =1*10-3/100=10-5

pH= -log (1*10-5)= 5

Change in pH= 7-5 =2

b)

NaOH is strong acid and ionizes completely

Hence [OH-] =0.051

pOH= -log (0.051)=1.3, pH= 14-pOH= 14-1.3= 12.7

Moles of NaOH in 100ml =0.051*100/1000 =0.0051

Moles of acid = 1*10-3 moles=0.001

Assuming the acid to be monoprotic acid, the reaction is

NaOH +HA------à NaA + H2O

1 mole of acid requires 1 mole of base

Moles of acid is limiting and excess NaOH=0.0051-0.001 =0.0041

Concentration of NaOH= 0.0041*1000/100 =0.041M

pOH= -log (0.0041)= 2.4, pH= 14-2.4= 11.6

Change in pH= 12.7-11.6= 1.1

c)Acetic acid is weak acid. Its ionization is represented as

CH3COOH-------à CH3COO- +H+

Ka= [CH3COO-] [H+]/ [CH3COOH]

Let x= [H+] = [CH3COO-]

Ka= x2/ (0.051-x)= 1.8*10-5

when solved using excel, x=0.00095

pH= 3.02

When 1*10-3 mmmol of acid is added, total moles of H+

1*10-3+0.00095=0.00195

Concentration of [H+] =0.00195*1000/100 = 0.0195M

pH= -log(0.0195)= 1.71

change in pH= 1.71-3.02= -1.31

d) CH3COO-+ H2O ------CH3COOH+OH-

Kb =10-14/Ka. For acetic acid Ka= 1.8*10-5

Kb= 10-14/1.8*10-5 =5.56*10-10

Kb= x2/ (0.051-x), where x =[OH-]

When solved using excel, [OH-] =0.000529

Concentration of acetic acid formed = 0.000529

pOH= -log (0.000529)=3.27

pH= 14-3.27= 10.73

when HCl is added

HCl reacts with CH3COONa as per the following equation

CH3COONa +HCl------à CH3COOH+ NaCl

CH3COOH is formed additionally

Moles of HCl =0.001, mole of CH3COONa= 0.051*100/1000 = 0.0051

Moles of CH3COOH is excess and excess by =0.0051-0.001= 0.0041

Concentration of sodium acetate = 0.0041*1000/100 =0.041M

Moles of CH3COOH= 0.001

Concentration of CH3COOH= 0.001*1000/100 +0.000529 =0.010529

pH= pka + log[ CH3COO-]/[CH3COOH]

pKa of acetic acid =4.72

pH= 4.74+ log [ 0.041/0.010529]=5.3

change in pH= 10.73-5.3= 5.23

e)

c)Acetic acid is weak acid. Its ionization is represented as

CH3COOH-------à CH3COO- +H+

Ka= [CH3COO-] [H+]/ [CH3COOH]

Let x= [H+] = [CH3COO-]

Ka= x2/ (0.051-x)= 1.8*10-5

when solved using excel, x=0.00095

pH= 3.02

for a mixture of CH3COOH and acetic acid having same concentration at 0.051M

when HCl is added , moles of HCl =0.001.

CH3COONa+HCl-----à CH3COOH + Nacl

Moles of CH3COOH= 100*0.051/100 =0.0051. HCl moles =0.001 and it is the limiting reactants

Moles of CH3COOH formed= 0.001

Moles of CH3COOH total =0.0041+0.001 =0.0051

Concentration of CH3COOH= 0.0051*1000/100 =0.051

Moles of sodium acetate= 0.0051-0.001= 0.0041

Concentration = 0.041*1000/100=0.041

pH= 4.74 +log (0.041/0.051)=4.64

Change in pH= 4.64-3.02= 1.62

f)0.51 M CH3COOH and 0.51 M CH3COONa

Ka= x2/(0.51-x)= 1.8*10-5

When solved x= 0.003, pH= 2.52

When HCl is added, CH3COONa reacts with HCl to form more CH3COOH

So more CH3COOH is formed and moles formed= 0.001( HCl is limiting

Moles of CH3COONa remaining =0.51*100/1000 -0.001=0.05

Concentration of CH3COONa= 0.05*1000/100 =0.5

Moles of CH3COOH= 0.51*100/1000 + 0.001=0.052

pH= 4.74+log (0.5/0.52)= 4.72

change in pH= 4.72-2.52= 2.2

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