Item 9 A 50.0 mL sample of 0.120 M HBr is titrated with 0.240 MNaOH. Calculate t

Question

Item 9
A 50.0 mL sample of 0.120 M HBr is titrated with 0.240 MNaOH. Calculate the pH after the addition of the following volumes of base.

Part A.)

0.0 mL

Express your answer using two decimal places.

Part B.)

20.0 mL

Express your answer using two decimal places.

Part C.)

24.9 mL

Express your answer using one decimal place.

Part D.)

25.0 mL

Express your answer using two decimal places.

Part E.)

25.1 mL

Express your answer using two decimal places.

Part F.)

40.0 mL

Express your answer using two decimal places.

Explanation / Answer

Answer – We are given, [HBr] = 0.120 M , volume = 50.0 mL , [NaOH] = 0.240 M

Part A) pH after adding 0.0 mL of base –

When there is no any base is added then there is only acid and HBr is strong acid, so

[HBr] = [H+] = 0.120 M

We know ,

pH = -log [H+]

     = - log 0.120

     = 0.921

Part B) pH after adding 20.0 mL of base-

We know when NaOH is strong base and HBr also strong acid, so when then added to each other there is neutralization reaction as follow –

HBr + NaOH -----> NaBr + H2O

Calculation of moles –

Moles of HBr = 0.120 M * 0.050 L = 0.006 moles

Moles of NaOH = 0.240 M *0.020 L = 0.0048 moles

So , moles of acid is excess

Remaining moles of acid after reaction = 0.006 moles – 0.0048 moles

                                                                = 0.0012 moles

Total volume = 50+20 = 70 mL

So, new concentration of H+

[H+] = 0.0012 moles / 0.070 L

        = 0.0171 M

So,

We know ,

pH = -log [H+]

     = - log 0.0171

     = 1.76

Part C) pH after adding 24.9 mL of base-

Calculation of moles –

Moles of HBr = 0.120 M * 0.050 L = 0.006 moles

Moles of NaOH = 0.240 M *0.0249 L = 0.00598 moles

So , moles of acid is excess

Remaining moles of acid after reaction = 0.006 moles – 0.00598 moles

                                                                = 2.4E-5 moles

Total volume = 50+24.9 = 74.9 mL

So, new concentration of H+

[H+] = 2.4E-5 moles / 0.0749 L

        = 0.00032 M

So,

We know,

pH = -log [H+]

     = - log 0.00032

     = 3.49

Part D) pH after adding 25 mL of base-

Calculation of moles –

Moles of HBr = 0.120 M * 0.050 L = 0.006 moles

Moles of NaOH = 0.240 M *0.025 L = 0.006 moles

So, both acid and base reacted to each other completely and there is formed neutral pH

So pH = 7.0

Part E) pH after adding 25.1 mL of base-

Calculation of moles –

Moles of HBr = 0.120 M * 0.050 L = 0.006 moles

Moles of NaOH = 0.240 M *0.0251 L = 0.006024 moles

So , moles of base is excess

Remaining moles of base after reaction = 0.006024 moles – 0.0060 moles

                                                                = 2.4E-5 moles

Total volume = 50+25.1 = 75.1 mL

So, new concentration of OH-

[OH-] = 2.4E-5 moles / 0.0751 L

          = 0.00032 M

So,

We know,

pOH = -log [OH-]

     = - log 0.00032

     = 3.49

pH = 14-pOH

      = 14 – 3.49

      = 10.5

Part F) pH after adding 40.0 mL of base-

Calculation of moles –

Moles of HBr = 0.120 M * 0.050 L = 0.006 moles

Moles of NaOH = 0.240 M *0.025 L = 0.006 moles

So, both acid and base reacted to each other completely and there is formed neutral pH

So pH = 7.0

Part E) pH after adding 25.1 mL of base-

Calculation of moles –

Moles of HBr = 0.120 M * 0.050 L = 0.006 moles

Moles of NaOH = 0.240 M *0.040 L = 0.0096 moles

So , moles of base is excess

Remaining moles of base after reaction = 0.0096 moles – 0.0060 moles

                                                                = 0.0036 moles

Total volume = 50+40 = 90 mL

So, new concentration of OH-

[OH-] = 0.0036 moles / 0.090 L

          = 0.04 M

So,

We know,

pOH = -log [OH-]

     = - log 0.04 M

     = 1.40

pH = 14-pOH

      = 14 – 1.40

      = 12.6

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