Question

Perform the following calculation:

A 1.275 g sample of CaBr₂ is dissolved in enough water to give 750. mL of solution. WHat is the calcium ion concertration in this solution? Make sure to write a reaction for the process of dissolving.

Answer 1

1st find the concentration of CaBr2

Molar mass of CaBr2,

MM = 1*MM(Ca) + 2*MM(Br)

= 1*40.08 + 2*79.9

= 199.88 g/mol

mass(CaBr2)= 1.275 g

number of mol of CaBr2,

n = mass of CaBr2/molar mass of CaBr2

=(1.275 g)/(199.88 g/mol)

= 6.379*10^-3 mol

volume , V = 750 mL

= 0.75 L

Molarity,

M = number of mol / volume in L

= 6.379*10^-3/0.75

= 8.51*10^-3 M

Now CaBr2 when dissolved dissociates as:

CaBr2       —>    Ca2+       +   2 Br-

So,

[Ca2+] = [CaBr2] = 8.51*10^-3 M

Answer: 8.51*10^-3 M