Question

a) In the laboratory you dissolve 14.5 g of chromium(II) bromide in a volumetric flask and add water to a total volume of 500. mL. What is the molarity of the solution?

b) In the laboratory, a student adds 56.1 mL of water to 16.1 mL of a 0.714 M hydroiodic acid solution. What is the concentration of the diluted solution? Assume the volumes are additive.

c) Calculate the volume of 0.220-M NaOH solution needed to completely neutralize 28.5 mL of a 0.860-M solution of the diprotic acid H₂C₂O₄.

Answer 1

a)

Molarity = mol / volume

mol = mass/MW = 14.5/ 211.80 = 0.06846 mol of CrBr2

V = 500 mL = 0.5 L

M = mol/V = 0.0684/.5 = 0.1368 M

b)

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

V1 = 16.1 mL M1 = 0.714

Vfinal = (V1 + V2) Mfinal =?

Vfinal = 16.1 + 56.1 = 72.2 mL

V1*M1 = V2*M2

16.1 *0.714 = 72.2 *M2

M2 = 16.1 *0.714/72.2  

M = 0.1592 mol per liter

c)

2H+ + 2OH- <--> 2H2O

mmol of acid = MV = 28.5*0.860 = 24.51 mmol of acid

1 mmol of acid = 2 mmol of H+

24.51 mmol of acid = 2*24.51 0 49.02 mmol of H+

ratio for neutralization is 1:1

mmol of bas e= mmol of acid

24.51 mmol of H+ = 24.51 mmol of OH-

[NaOH] = 0.22 M

V = mol/M = 24.51 /0.22

V = 111.409 mL