A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

Respuesta :

Answer: The p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

  • Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, [tex]M_1=(2\times 0.0031)=0.0062M[/tex]

  • Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, [tex]M_2=(2\times 0.0042)=0.0084M[/tex]

To calculate the concentration of nitrate ions in the solution, we use the equation:

[tex]M=\frac{M_1V_1+M_2V_2}{V_1+V_2}[/tex]

Putting values in above equation, we get:

[tex]M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M[/tex]

Calculating the p-function of zinc ions and nitrate ions in the solution:

  • For zinc ions:

[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}][/tex]

[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51[/tex]

  • For nitrate ions:

[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}][/tex]

[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14[/tex]

Hence, the p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.