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Question: We define the p value of a term (say x) as \({{p}_{x}}=-\log x\). Calculate the value of 1000p for \...

We define the p value of a term (say x) as px=−log⁡x{{p}_{x}}=-\log x. Calculate the value of 1000p for NO3−NO_{3}^{-} in a solution that is 4.78×10−2M4.78\times {{10}^{-2}}M in Cu(NO3)2Cu{{\left( N{{O}_{3}} \right)}_{2}} and 0.104M in Zn(NO3)2Zn{{\left( N{{O}_{3}} \right)}_{2}}.

Explanation

Solution

It is given px=−log⁡x{{p}_{x}}=-\log x where ‘x’ is the concentration. We need to calculate the 1000p value for nitrate ions, NO3−NO_{3}^{-} in the two given solutions of copper nitrate and zinc nitrate. Start by writing the dissociation reaction of copper nitrate and zinc nitrate and then from the concentration of solutions given, find concentration of nitrate ions and then substitute the value in the given formula to get p value. For 1000p, just multiply p value by 1000.

Complete answer:
- Let’s start by writing the dissociation reaction for Cu(NO3)2Cu{{\left( N{{O}_{3}} \right)}_{2}}
Cu(NO3)2⇌Cu2++2NO3−Cu{{\left( N{{O}_{3}} \right)}_{2}}\rightleftharpoons C{{u}^{2+}}+2NO_{3}^{-}
- Copper nitrate dissociates to give one cupric ion and two nitrate ions.
- We know that concentration of a species in an electrolyte depends on the number of moles of that species present in the solution. Therefore, concentration of nitrate ions will be double the concentration of copper nitrate.
- Therefore, conc. of NO3−NO_{3}^{-} ions is 2×4.78×10−2M2\times 4.78\times {{10}^{-2}}M.
- Now, we have the equation, px=−log⁡x{{p}_{x}}=-\log x where x is the concentration.
- Therefore, pNO3−=−log⁡[NO3−]{{p}_{NO_{3}^{-}}}=-\log \left[ NO_{3}^{-} \right]
- Substituting the value of conc. of NO3−NO_{3}^{-} we get,

& {{p}_{NO_{3}^{-}}}=-\log \left[ NO_{3}^{-} \right] \\\ & =-\log \left( 2\times 4.78\times {{10}^{-2}} \right) \\\ & {{p}_{NO_{3}^{-}}}=1.02 \end{aligned}$$ \- Therefore, ${{p}_{NO_{3}^{-}}}=1.02$ and hence, $1000{{p}_{NO_{3}^{-}}}=1020$ \- Similarly, zinc nitrate, $Zn{{\left( N{{O}_{3}} \right)}_{2}}$ dissociates into one zinc (II) ion and two nitrate ions. $$Zn{{\left( N{{O}_{3}} \right)}_{2}}\rightleftharpoons Z{{n}^{2+}}+2NO_{3}^{-}$$ \- Therefore, conc. of $NO_{3}^{-}$ ions is $2\times 0.104M$. \- Substituting the value of conc. of $NO_{3}^{-}$ ions in equation, ${{p}_{NO_{3}^{-}}}=-\log \left[ NO_{3}^{-} \right]$, we get, $$\begin{aligned} & {{p}_{NO_{3}^{-}}}=-\log \left[ NO_{3}^{-} \right] \\\ & =-\log \left( 2\times 0.104 \right) \\\ & {{p}_{NO_{3}^{-}}}=0.682 \end{aligned}$$ \- Therefore, ${{p}_{NO_{3}^{-}}}=0.682$ and hence, $1000{{p}_{NO_{3}^{-}}}=682$ \- Therefore, the value of 1000p for $NO_{3}^{-}$ in a $Cu{{\left( N{{O}_{3}} \right)}_{2}}$ solution and a $Zn{{\left( N{{O}_{3}} \right)}_{2}}$ solution is 1020 and 682 respectively. **Note:** Remember concentration of an ion in an electrolytic solution depends on the number of moles of that species present in the solution. p-value indicates the potential or strength of a species present in the solution. p-value of an ion is the negative logarithm of concentration of that specific ion present in the solution.