Solution Found!
How much 0.100 M HCl is required to completely neutralize 20.0 mL of 0.250 M NaOH?
Chapter 14, Problem 89P(choose chapter or problem)
What substance could you add to each solution to make it a buffer solution?
(a) \(\mathrm{0.100~ M ~NaC_2H_3O_2}\)
(b) \(\mathrm{0.500~ M~ H_3PO_4}\)
(c) \(\mathrm{0.200~ M~ HCHO_2}\)
Equation Transcription:
Text Transcription:
0.100 M NaC_2H_3O_2
0.500 M H_3PO_4
0.200 M HCHO_2
Questions & Answers
QUESTION:
What substance could you add to each solution to make it a buffer solution?
(a) \(\mathrm{0.100~ M ~NaC_2H_3O_2}\)
(b) \(\mathrm{0.500~ M~ H_3PO_4}\)
(c) \(\mathrm{0.200~ M~ HCHO_2}\)
Equation Transcription:
Text Transcription:
0.100 M NaC_2H_3O_2
0.500 M H_3PO_4
0.200 M HCHO_2
ANSWER:
Step 1:
How much 0.100 M HCl is required to completely neutralize 20.0 mL of 0.250 M NaOH?
Given :
Molarity of HCl(Hydrochloric acid)(M1) = 0.100 M
Molarity of NaOH(sodium hydroxide) (M2) = 0.250 M
Volume of NaOH(V2)= 20.0 mL
Volume of HCl to neutralize NaOH (V1) = ?
Let’s write the chemical reaction for the given scenario :
HCl +NaOHNaCl+H20
We know the formula to calculate the molarity of the diluted solution, :
M1V1 = M2V2
Where, M1 and V1 = Molarity and volume of initial concentrated solution.
M2 and V2 = Molarity and volume of final diluted solution.
Using the above formula we calculate the volume of HCl to neutralize NaOH.
We have to calculate V1, hence the formula becomes :
V1 =
Volume of NaOH is in mL, so let’s convert to L :
1 L = 1000 mL
Therefore, 20.0 mL in L will be :
= 20 mL
= 0.020 L