Determine the pH during the titration of 28.5 mL of 0.232 M HNO3 by 0.232 M NaOH at the following points: (a) Before the addition of any NaOH (b) After the addition of 14.3 mL of NaOH (c) At the equivalence point (d) After adding 35.3 mL of NaOH
(a) Before the addition of any NaOH
(b) After the addition of 14.3 mL of NaOH
(c) At the equivalence point
(d) After adding 35.3 mL of NaOH
Given,
The titration of 28.5 mL of 0.232 M HNO3 by 0.232 M NaOH .
• First, let's take a look at the acid-base reaction. The equation for this one is:
HNO3 + NaOH --> H2O + NaNO3
We also need to know the equation for pH, which is:
pH = -log[H+].
(1) Before the addition of any sodium hydroxide, you know that you have a solution that is 0.232M [HNO3]
Since [H +] = [NO3-] = [HNO3], you know that [H+] = 0.232 .
• pH=-log(0.232) = 0.635 .
(2) Now, you need to know how many moles of your acid were neutralized by the NaOH.
Numbers of moles of NaOH
= (0.0143 L * 0.232 mol .L-1)
= 0.0033176 moles
To find out how many moles of HCl you started with:
No. Of moles of HNO3
= (0.0285 L * 0.232 mol.L-1)
= 0.006612 moles
No. Of excess moles of HNO3
= (0.006612 - 0.0033176) moles
= 0.0032944 moles .
Total volume = ( 0.0285 + 0.0143) L
= 0.0428 L .
Concentration of H+ ,
[H+] = (0.0032944/0.0428) M = 0.07697 M
• pH = - log (0.07697) = 1.11 .
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