Calculate the pH of the solution before and after the addition of NaOH.?

Question

A 25.0 mL sample of 0.100 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution before the addition of NaOH and after the addition of (a) 25.0 ML (b) 30.0 mL of NaOH.

Answer 1

ANSWERS:

Initially:
pH = 1

@ 25.0 mL of 0.100 M NaOH:
pH = 7

@ 30.0 mL of 0.100 M NaOH:
pH = 11.96


THE SOLUTIONS:

During the titration of a strong acid with a strong base:

The initial pH of the solution, that is, without adding the base titrant, is mainly due to the concentration of the analyte.

Hence,

@ Volume of 0.100 M NaOH = 0,

the [H+] = [HNO3]

.:. [H+] = 0.100 M

pH = -log[H+]

.:. pH = 1



(a) Volume of 0.100 M NaOH = 25.0 mL,

25.0 mL * 0.100 M HNO3 = 2.5 mmol HNO3
25.0 mL * 0.100 M NaOH = 2.5 mmol NaOH = 2.5 mmol HNO3
------------------------------------------------------------------------------------------------------------------------
.:. the difference is 0. This must be the equivalence point.

At the equivalence point of the titration of a strong acid vs. a strong base, the pH is always equal to 7. Hence,

.:. pH = 7



(b) Volume of 0.100 M NaOH = 30.0 mL,

*This must be the POST-EQUIVALENCE point, therefore, the change in the pH of the solution must be due to the excess OH- ions present in the solution, so...

30.0 mL * 0.100 M NaOH = 3.0 mmol NaOH
25.0 mL * 0.100 M HNO3 = 2.5 mmol HNO3 = 2.5 mmol NaOH
------------------------------------------------------------------------------------------------------------------------
.:. the difference is 0.5 mmol NaOH (in excess)


To solve for the concentration, we must find the TOTAL VOLUME of the solution:

TOTAL VOLUME = 25.0 mL + 30.0 mL

.:. TOTAL VOLUME = 55.0 mL

[OH-] = 0.5 mmol NaOH / 55.0 mL

.:. [OH-] = 0.0091 M

pOH = -log[OH-]

.:. pOH = 2.04

pH = 14 - pOH

.:. pH = 11.96

Answer 2

before 0.1M HNO3 pH =log1/.1= = log 10 =1
with 25ml equivalent point pure salt ph =7

with 30ml here are 5mL NaOH in 130ml
concentration OH- 0.1*5/130 =0.00384

pOH =log 260 = 2.42

pH=14-pO= 14-2.42= 11.58

Answer 3

Unless you are a "Dr." of a social science or literature, you should have a decent stockpile of textbooks that tackle a range of chemistry subjects. I would start with your first college chemistry textbook. It will explain molarity, concentration, pH, stoichiamatry. If I recall my chemistry classes correctly this would have been about as easy as the questions would ever be. Do your homework, go to class, and you won't have to trust self righteous, judgemental, know-nothing-know-it-alls on Y! answers.

Though you could be trying to whip up a batch of meth. In that case you might want to be a little more discreet.

It you are a Dr. of a field not related to chemistry then don't quit your day job or do your own homework.