Strong Acid Strong Base Titrations

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The titration of 50.0mL of 0.150M HCl with 0.150M NaOH is carried out in a chemistry laboratory. Calculate the pH of the solution after these volumes of the titrant have been added. Use these results to plot the titration curve.

pH	Vol. NaOH
	0.0mL
	25.0mL
	49.9mL
	50.0mL
	50.1mL
	75.0ml

0.0ml of NaOH 50.0mL 0.150M HCl

pH=-log0.150M

pH= 0.82

25.0ml of NaOH to 50.0mL of HCl

Moles H⁺=50.0mL x 0.150M=7.50mmol

Moles OH^-=25.0mL x 0.150M=3.75mmol

	H⁺ +	OH^-	è H₂O
I	7.50mmol	3.75mmol
C	-3.75mmol	-3.75mmol
E	3.75mmol	0

[H⁺]=	3.75mmol	=0.05M
	25.0mL +50.0mL

pH=-log[H⁺]

pH=-log[0.05M]=0.82

49.9ml of NaOH to 50.0mL of HCl

Moles H⁺=50.0mL x 0.150M=7.50mmol

Moles OH^-=49.9mL x 0.150M=7.485mmol

	H⁺ +	OH^-	è H₂O
I	7.50mmol	7.485mmol
C	-7.485mmol	-7.485mmol
E	0.015mmol	0

[H⁺]=	0.015mmol	=0.000 15M
	49.9mL +50.0mL

pH=-log[H⁺]

pH=-log[0.000 15M]=3.82

50.0ml of NaOH to 50.0mL of HCl

Moles H⁺=50.0mL x 0.150M=7.50mmol

Moles OH^-=50.0mL x 0.150M=7.50mmol

	H⁺ +	OH^-	è H₂O
I	7.50mmol	7.50mmol
C	-7.50mmol	-7.50mmol
E	0	0

neutralization pH=7

50.1ml of NaOH to 50.0mL of HCl

Moles H⁺=50.0mL x 0.150M=7.50mmol

Moles OH^-=50.1mL x 0.150M=7.515mmol

	H⁺ +	OH^-	è H₂O
I	7.50mmol	7.515mmol
C	-7.50mmol	-7.50mmol
E	0	0.015mmol

[OH^- ]=	0.015mmol	=0.000 15M
	50.1mL +50.0mL

pOH=-log[OH^-]

pOH=-log[0.000 15M]=3.82

pH=14-3.82=10.18

75.0ml of NaOH to 50.0mL of HCl

Moles H⁺=50.0mL x 0.150M=7.50mmol

Moles OH^-=75.0mL x 0.150M=11.25mmol

	H⁺ +	OH^-	è H₂O
I	7.50mmol	11.25mmol
C	-7.50mmol	-7.50mmol
E	0	3.75mmol

[OH^- ]=	3.75mmol	=0.030M
	75.0mL +50.0mL

pOH=-log[OH^-]

pOH=-log[0.030M]=1.52

pH=14-1.52=12.48

pH	Vol. NaOH
0.82	0.0mL
1.30	25.0mL
3.82	49.9mL
7.00	50.0mL
10.18	50.1mL
12.48	75.0ml

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