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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 _________

User Jolo
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1 Answer

3 votes

Answer:


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

User Mabahamo
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