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Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.
a. 0.100 mole of Ca(NO3)2 in 100.0 mL of solution.
b. 2.5 moles of Na2SO4 in 1.25 L of solution.
c. 5.00 g of NH4Cl in 500.0 mL of solution.
d. 1.00 g K3PO4 in 250.0 mL of solution.

Answer :

anfabba15

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

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