1. In the titration of 50.00 mL of 0.100 M HCI with 0.100 M NaOH, calculate the
ID: 1039216 • Letter: 1
Question
1. In the titration of 50.00 mL of 0.100 M HCI with 0.100 M NaOH, calculate the pH after (a) 0.00 ml, (b) 10.00 ml (c) 25.00 ml, and (d) 50.00 ml of NaOH have been added (at the equivalent point),. Choose a suitable indicator for this titration In the titration of 50.00 mL of 0.100 M HF with 0.100 M NaOH, calculate the pH after (a) 0.00 ml, (b) 10.00 ml(buffer region) (c) 25.00 ml, of NaOH have been added (1/2 the equivalent point) (d) 50.00 ml of NaOH have been added(equivalent point). Choose a suitable indicator for this titration. (Ka = 6.8 x 10-4 ) 2.Explanation / Answer
(1)
(a)
After 0 mL base has been added:
pH = -log(0.1) = 1
(b)
After 10 mL base is added:
Moles of acid left = (0.05*0.1 - 0.01*0.1) = 0.004
Total volume = 60 mL
So, [H+] = 0.004/0.06 = 0.067
So,
pH = -log(0.067) = 1.174
(c)
After 25 mL base is added:
Moles of acid left = (0.05*0.1 - 0.025*0.1) = 0.0025
Total volume = 75 mL
So, [H+] = 0.0025/0.075 = 0.033
So,
pH = -log(0.033) = 1.48
(d)
After 50 mL base is added:
Moles of acid left = (0.05*0.1 - 0.05*0.1) = 0
So at this point the pH is equal to 7, because this is the equivalence point.
Hope this helps !
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