Please help me with those question. 1. Calculate the pH during the titration of

Question

Please help me with those question.

1. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid:

7.00 mL

pH= ?

29.70 mL

pH= ?

38.00 mL

pH= ?

2. Calculate the pH during the titration of 40.00 mL of 0.1000 M HCI with 0.1000 M NaOH solution after the following additions of base:

20.00 mL

pH=

39.30 mL

pH=

48.00 mL

pH=

3. An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of 0.400 M HCIO and 0.450 M NaCIO. (KA for HCIO=2.9x10^-8)

pH=

4. Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka=1.54x10^-5), with 0.1000 M NaOH solution after the following additions of titrant.

14.00 ML:

pH= ?

20.50 mL:

pH= ?

30.00 ML:

pH= ?

Explanation / Answer

To solve this problem we have to find the moles of every specie, then we have to compare them and see which one is in excess, for example if you have 2 moles of acid and 1 mole of base then the acid will neutralize the base completely the base and you will have 1 mole of acid free in your solution this will dictate the ph of your solution.

First we calculate the number of moles in the KOH solution (30 ml, 0.1M) 30 ml = 0.03 L

moles of KOH = Molarity * Volume = 0.03 L * 0.1 M = 0.003 moles

Now we calculate the number of HBr moles for first part ( 7 ml, 0.1M)

moles of HBr = 0.007 * 0.1 = 0.0007 moles

Now we compare the number of moles KOH vs HBr

0.003 moles of base vs 0.0007 moles of acid. We can see that the are more moles of base than the acid ones.

0.003 - 0.0007 = 0.0023 moles of base are remaining

Now we calculate the total volume available which is 30 ml + 7 ml = 37 ml = 0.037 L

Concentration of OH = 0.0023 moles / 0.037 L = 0.0621 M

POH = -log (0.0621) = 1.2

PH = 12.8

The addition of 29.7 ml is analyzed the same way:

You find that there are ( 0.0297 L * 0.1 M ) = 0.00297 moles of acid

The ammount of base is greater than the ammount of acid so there are 0.00003 moles of base

Concentration of OH = 0.0003 / 0.0597 = 0.0005025 M

POH = 3.29

PH = 14-3.29 = 10.7

For part C we have more moles of acid than base

moles of acid = 0.038 L * 0.1 M = 0.0038 moles

Total volume = 0.068 L (30 + 38)

in this case there is more acid than base so:

0.0038 - 0.003 = 0.0008 moles of acid remaining

concentration of H = 0.0008 / 0.068 = 0.011764 M

PH = -log (0.011764) = 1.92

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