Question

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
m_{ca^{2+}} = m
m_{no₃^-} = m
b. 7.0 moles of na₂so₄ in 2.50 l of solution
m_{na^+} = m
m_{so₄^{2-}} = m
c. 3.10 g of nh₄cl in 890.0 ml of solution
m_{nh₄^+} = m
m_{cl^-} = m
d. 8.10 g of k₃po₄ in 490.0 ml of solution
m_{k^+} = m
m_{po₄^{3-}} = m

Explanation:

Step1: Calculate molarity formula

Molarity \(M=\frac{n}{V}\) (where \(n\) is the number of moles and \(V\) is the volume in liters)

Step2: For part a

• Given \(n_{Ca(NO_3)_2}=0.100\space mol\), \(V = 100.0\space mL=0.1000\space L\)
• \(M_{Ca(NO_3)_2}=\frac{0.100\space mol}{0.1000\space L}=1.00\space M\)
• Since \(Ca(NO_3)_2

ightarrow Ca^{2 +}+2NO_3^-\), \(M_{Ca^{2+}} = M_{Ca(NO_3)_2}=1.00\space M\)

• \(M_{NO_3^-}=2\times M_{Ca(NO_3)_2}=2\times1.00\space M = 2.00\space M\)

Step3: For part b

• Given \(n_{Na_2SO_4}=7.0\space mol\), \(V = 2.50\space L\)
• \(M_{Na_2SO_4}=\frac{7.0\space mol}{2.50\space L}=2.8\space M\)
• Since \(Na_2SO_4

ightarrow 2Na^{+}+SO_4^{2 -}\), \(M_{Na^{+}}=2\times M_{Na_2SO_4}=2\times2.8\space M = 5.6\space M\)

• \(M_{SO_4^{2-}}=M_{Na_2SO_4}=2.8\space M\)

Step4: For part c

• Molar mass of \(NH_4Cl\): \(M = 14.01+4\times1.008 + 35.45=53.49\space g/mol\)
• \(n_{NH_4Cl}=\frac{3.10\space g}{53.49\space g/mol}\approx0.05796\space mol\), \(V = 890.0\space mL = 0.8900\space L\)
• \(M_{NH_4Cl}=\frac{0.05796\space mol}{0.8900\space L}\approx0.0651\space M\)
• Since \(NH_4Cl

ightarrow NH_4^{+}+Cl^-\), \(M_{NH_4^{+}}=M_{NH_4Cl}\approx0.0651\space M\)

• \(M_{Cl^-}=M_{NH_4Cl}\approx0.0651\space M\)

Step5: For part d

• Molar mass of \(K_3PO_4\): \(M=(3\times39.10)+30.97+(4\times16.00)=212.27\space g/mol\)
• \(n_{K_3PO_4}=\frac{8.10\space g}{212.27\space g/mol}\approx0.0382\space mol\), \(V = 490.0\space mL=0.4900\space L\)
• \(M_{K_3PO_4}=\frac{0.0382\space mol}{0.4900\space L}\approx0.0780\space M\)
• Since \(K_3PO_4

ightarrow 3K^{+}+PO_4^{3 -}\), \(M_{K^{+}}=3\times M_{K_3PO_4}=3\times0.0780\space M = 0.234\space M\)

• \(M_{PO_4^{3-}}=M_{K_3PO_4}\approx0.0780\space M\)

Answer:

a. \(M_{Ca^{2+}} = 1.00\space M\), \(M_{NO_3^-}=2.00\space M\)
b. \(M_{Na^{+}} = 5.6\space M\), \(M_{SO_4^{2-}}=2.8\space M\)
c. \(M_{NH_4^{+}}\approx0.0651\space M\), \(M_{Cl^-}\approx0.0651\space M\)
d. \(M_{K^{+}} = 0.234\space M\), \(M_{PO_4^{3-}}\approx0.0780\space M\)