Question

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 _____.

Answer 1

Answer: The p-function of $Zn^{2+}$ and $NO_3^{-}$ 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, $M_1=(2\times 0.0031)=0.0062M$

• 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, $M_2=(2\times 0.0042)=0.0084M$

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

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

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

• For zinc ions:

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

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

• For nitrate ions:

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

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

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