Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.

Answer 1

Answer:

a. Ca2+  1M ; NO3-  2M

b. Na+  4M ;  SO4-2  2M

c.  NH4+ , Cl-  0,186M  

d. K+ 0,056M ; PO4-3 0,0188 M

Explanation:

Here are the ionization equations

Ca(NO3)2 ----> Ca2+  +  2NO3-

Na2SO4 ------> 2Na+   +  SO4-2

NH4Cl ------> NH4+  +  Cl-

K3PO4 -----> 3K+   +   PO4-3

Concentrations must be in Molarity (moles/L)

Ca(NO3)2 0,1 mole in 100mL

100mL = 0,1 L

Molarity Ca(NO3)2  1M  (0,1 /0,1)

Ca(NO3)2 ----> Ca2+  +  2NO3-

1 M                      1M            2M

Na2SO4 2,5 moles in 1,25L

Molarity Na2SO4 2,5 moles /1,25L = 2M

Na2SO4 ------> 2Na+   +  SO4-2

2M                     4M             2M

5.00 g of NH4Cl in 500mL

We have to resort to the molar mass

Let's make the rule of three

500mL ______ 5g

1000mL _____ (1000mL . 5g) / 500mL = 10 g

(These are the grams in 1L, so with the molar mass, we get the moles)

Mass / Molar mass = moles --> 10g / 53,5 g/m = 0.186 moles

NH4Cl ------> NH4+  +  Cl-

0,186 M      0,186M  +  0,186M

1.00 g K3PO4 in 250.0 mL

(The same as before)

250 mL ________1 g

1000 mL _______ (1000 mL . 1 g) /250mL = 4 g

These are the grams in 1L, so with the molar mass, we get the moles

Mass / Molar mass = moles -->  4g /212,26 g/m = 0,0188 moles

K3PO4 -----> 3K+   +   PO4-3

0,0188 M      0,056M   +  0,0188 M