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

Answer:

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

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

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

$M_B = \frac{M_A * V_A }{V_B}$

This gives:

$M_B = \frac{0.016M *20mL}{18.3mL}$

$M_B = \frac{0.016M *20}{18.3}$

$M_B = \frac{0.32M}{18.3}$

$M_B = 0.0175M$

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