Question

A sample of 180.0 mL of 1.65M HNO3 is titrated with 0.90M KOH. Fill in the following table for each different point (a, b, c, etc….) during the titration. Also graph the titration curve with pH on the y­axis and Vbase on the x­axis (Make sure all work is attached for credit)

Point |Vbase (mL)| molbase added| mol acid remain| molbase remain| [OH ­ ]| [H3O + ]| pH|

a            0.00

b           75.00

c           200.25

d           275.50

e            Veq=

f            400

Answer 1

The equivalence point EP is at V:

V x 0.90 mol/L = 0.180 L x 1.65 mol/L    

V= 0.33 L = 330 mL KOH

For the Excel Table

Initial acid quantity is

0.180 L x 1.65 mol/L = 0.297 mol

Base, mol is    Vb/1000 x 0.9 M

Acid remain, mol     0.297-Base

Base remain, mol     Base – 0.297

point	Vb, mL	base, mol	acid remain,mol	base remain, mol	pH	pOH	[H+]	[HO-]
a	0	0	0.297	0	0.53	13.47	0.297	3.37E-14
b	75	0.068	0.230	0	0.64	13.36	0.230	4.36E-14
c	200.25	0.180	0.117	0	0.93	13.07	0.117	8.56E-14
d	275.5	0.248	0.049	0	1.31	12.69	0.049	2.04E-13
e EP	330	0.297	0	0	7	7	1.00E-07	1.00E-07
f	400	0.36	0	0.063	12.80	1.20	1.59E-13	0.063

Do a graph by hand using pH versus Vb columns.