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Trial 12 1.Volume of Acid used ( mL) 2020 2.Molarity of acid used (M) 0.0160.016 3.Initial volume reading of base buret 018.6 4.Final volume reading of base used 18.636.9 5.Volume of base used (4-3, mL) _________ _________ 6.Molarity of base (1*2/5, M) _________ _________ 7.Average molarity of base _________

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MrRoyal

Answer:

[tex]V_B = 18.3mL[/tex] -- Volume of base used

[tex]M_B = 0.0175M[/tex] --- Molarity of base

Explanation:

Given

[tex]V_A = 20mL[/tex] -- Volume of acid used

[tex]V_B_1 = 18.6mL[/tex] --- Buret Initial reading

[tex]V_B_2 = 36.9mL[/tex] --- Buret Final reading

[tex]M_A = 0.016M[/tex] --- Molarity of the acid

Solving (a): Volume of base used (VB)

This is calculated by subtracting the initial reading from the final reading of the base buret.

i.e.

[tex]V_B = V_B_2 - V_B_1[/tex]

[tex]V_B = 36.9mL - 18.6mL[/tex]

[tex]V_B = 18.3mL[/tex]

Solving (b): Molarity base (MB)

This is calculated using:

[tex]M_A * V_A = M_B * V_B[/tex]

Make MB the subject

[tex]M_B = \frac{M_A * V_A }{V_B}[/tex]

This gives:

[tex]M_B = \frac{0.016M *20mL}{18.3mL}[/tex]

[tex]M_B = \frac{0.016M *20}{18.3}[/tex]

[tex]M_B = \frac{0.32M}{18.3}[/tex]

[tex]M_B = 0.0175M[/tex]

Solving (c): There is no such thing as average molarity

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