Question

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 _________

Answer 1

`V_B = 18.3mL` -- Volume of base used

`M_B = 0.0175M` --- Molarity of base

Step-by-step explanation:

Given

`V_A = 20mL` -- Volume of acid used

`V_B_1 = 18.6mL` --- Buret Initial reading

`V_B_2 = 36.9mL` --- Buret Final reading

`M_A = 0.016M` --- 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.

`V_B = V_B_2 - V_B_1`

`V_B = 36.9mL - 18.6mL`

`V_B = 18.3mL`

Solving (b): Molarity base (MB)

This is calculated using:

`M_A * V_A = M_B * V_B`

Make MB the subject

`M_B = (M_A * V_A )/(V_B)`

This gives:

`M_B = (0.016M *20mL)/(18.3mL)`

`M_B = (0.016M *20)/(18.3)`

`M_B = (0.32M)/(18.3)`

`M_B = 0.0175M`

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