Question

Starting with the 0.1250M Cu(NH3)+2 as the stock (concentration) calculate the following concentrations of the dilute solutions showing work:

15mL stock diluted to 20mL

10.0mL of stock solution diluted to 20.0mL

5.0mL of the stock solution diluted to 20.0mL

Answer 1

Concentration of the stock solution of Cu(NH₃)²⁺ = 0.1250M

During diution we only add water. Hence the moles of Cu(NH₃)²⁺ before and after dilution remains constant.

(a). 15mL 0.1250M Cu(NH₃)²⁺ solution stock diluted to 20mL

moles of Cu(NH₃)²⁺ before dilution = MxV = 0.1250Mx15x10^-3 = 1.875x10^-3 moles

moles of Cu(NH₃)²⁺ after dilution = MxV = Mx20x10^-3 moles

moles of Cu(NH₃)²⁺ before dilution = moles of Cu(NH₃)²⁺ after dilution

1.875x10^-3 moles = Mx20x10^-3 moles

M = 1.875x10^-3 moles/20x10^-3 moles = 0.09375 M (answer)

(b) 10.0mL of stock solution diluted to 20.0mL

We can calculate this in a similar manner as we did in (a). But here the initial volume is 10mL

moles of Cu(NH₃)²⁺ before dilution = MxV = 0.1250Mx10x10^-3 = 1.250x10^-3 mol

moles of Cu(NH₃)²⁺ after dilution = MxV = Mx20x10^-3 mol

1.250x10^-3 mol =Mx20x10^-3 mol

M = 1.250x10^-3 mol/20x10^-3 mol = 0.0625M (answer)

(c) 5.0mL of the stock solution diluted to 20.0mL

similarly

moles of Cu(NH₃)²⁺ before dilution = MxV = 0.1250Mx5x10^-3 = 0.0625x10^-3 mol

moles of Cu(NH₃)²⁺ after dilution = MxV = Mx20x10^-3 mol

0.06250x10^-3 mol =Mx20x10^-3 mol

M = 0.03125 M (answer)