Question

A25.0 mL sample of 0.100 M NaOH is titrated with a 0.100 M HNO3 solution. Calculate the pH after the addition of 0.0, 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9, 25.0, 25.1, 26.0, 28.0, and 30.0 of the HNO3.

Answer 1

Moles of NaOH= 0.1*25/100= 0.025M

moles of HNO3 =0 ml

Since NaOH and HNO3 completely dissociate

POH= -log (0.25)= 1.6 and PH= 14-1.6= 12.4

when 4 ml of NH3 is added = (4/1000)*0.1 =0.004 moles

both NaOH and HNO3 are strong base and acid and ionize completely and react in 1:1 ratio.

Hence moles of NaOH is excess and this exess moles are =0.0025-0.0004=0.0021

volume of the solution after mixing =25+4 =29 ml

Molarity of NaOH= 0.0021*1000/29=0.0724, POH= -log (0.0724)= 1.14, PH= 14-1.14=12.86

Moles of HNO3 in 8ml =0.1*8/1000=0.0008 moles

Moles of NaOH unreacted =0.0025- 0.0008= 0.0017 Moles

Molarity = 0.0017*1000/(25+8)= 0.051 , PH= 14- (-log0.051)= 12.81

12.5 ml addition leads to HNO3 moles =(12.5/1000)*0.1=0.00125 Moles

Moles of NaOH unreacted= 0.0025-0.00125= 0.00125,

Molairty = 0.00125*100/37.5= 0.0333 Ph= 14-(-log(0.03333)= 12.52

20 ml addition gives rise to 0.0025- (20*0.1/1000)=0.0005 moles of NaOH, molarity= 0.0005/0.045 L = 0.01111

pH= 14- (-log(0.0111) =12.05

24 ml give rise to excess of NaOH = 0.0025- (24/1000)*0.1 =0.0001 , molarity = 0.0001/49=0.002041

pH= 14- (-log0.002041)= 11.3098

24.9 ml give rise to excess NaOH = 0.0025- 0.00249 =10-5moles, molaroty= 10-5/0.0449 = 0.0002

pH= 14- (-log (0.0002)= 10.3

when 25 ml is added, the solution becomes neutrial and the pH =7

at 25.1ml of HNO3 , molels added = (25.1/1000)*0.1=0.00251

Moles ofHNO3 is excess by =0.00251-0.0025= 1*10-5

Molarity of HNO3= 10-5*1000/50.1= 0.002 ,pH= -log (0.002)= 3.6999

When 26 ml of HNO3 is added, excess HNO3= 26*0.1/1000-0.0025= 1*10-4 molarity = 1*10-4*1000/(25+26)= 0.001961 M

pH =-log (0.001961)= 2.7

When 28ml is added excess HNO3= 28*0.1/1000-0.0025= 0.003 and molairty = 0.003*1000/53= .00566M

pH= -log (0.00566)= 2.24

when 30 ml is added excess HNO3= 30*0.1/1000-0.0025= 0.0005 ,molairty= 0.0005*1000/55= 0.0090

pH= -log (0.0090)= 2.04

When 8 ml is added, Moles of NaOH remaining = .025-0.008=0.017

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