when 0.0296 g of co(no₃)₂ is dissolved in 0.440 l of 0.365 m koh, what are co²⁺, co(oh)₄²⁻, and oh⁻? note…

Explanation:

Step1: Calculate moles of Co(NO₃)₂

First, find the molar - mass of Co(NO₃)₂. The molar mass of Co = 58.93 g/mol, N = 14.01 g/mol, and O = 16.00 g/mol. The molar mass of Co(NO₃)₂ is (M = 58.93+2\times(14.01 + 3\times16.00)=182.94) g/mol. The moles of Co(NO₃)₂, (n=\frac{m}{M}=\frac{0.0296\ g}{182.94\ g/mol}=1.62\times10^{-4}\ mol). The initial concentration of (Co^{2 +}), (C_{Co^{2+},initial}=\frac{n}{V}=\frac{1.62\times10^{-4}\ mol}{0.440\ L}=3.68\times10^{-4}\ M).

Step2: Consider the reaction with KOH

The reaction between (Co^{2+}) and (OH^-) to form (Co(OH)_4^{2 -}) is (Co^{2+}+4OH^-\rightleftharpoons Co(OH)4^{2 -}), and the formation constant (K_f) (which needs to be looked - up from the table). The concentration of (OH^-) initially is (C{OH^-}=0.365\ M). Since (K_f) for (Co(OH)_4^{2 -}) is very large, we assume that almost all of the (Co^{2+}) reacts to form (Co(OH)_4^{2 -}). Let (x) be the concentration of (Co^{2+}) at equilibrium. Then the concentration of (Co(OH)4^{2 -}) is approximately (3.68\times10^{-4}\ M) (assuming complete reaction of (Co^{2+}) initially), and the concentration of (OH^-) at equilibrium is (C{OH^-}=0.365 - 4\times3.68\times10^{-4}\approx0.365\ M) (because (4\times3.68\times10^{-4}\ll0.365)). Using the formation - constant expression (K_f=\frac{[Co(OH)_4^{2 -}]}{[Co^{2+}][OH^-]^4}), we can solve for ([Co^{2+}]). Rearranging for ([Co^{2+}]) gives ([Co^{2+}]=\frac{[Co(OH)_4^{2 -}]}{K_f[OH^-]^4}). Let's assume (K_f = 5\times10^{9}) (a typical value for (Co(OH)_4^{2 -})). Substituting ([Co(OH)_4^{2 -}]=3.68\times10^{-4}\ M) and ([OH^-]=0.365\ M) into the equation: [ \begin{align*} [Co^{2+}]&=\frac{3.68\times10^{-4}}{5\times10^{9}\times(0.365)^4}\ &=\frac{3.68\times10^{-4}}{5\times10^{9}\times0.01897}\ &=\frac{3.68\times10^{-4}}{9.485\times10^{7}}\ & = 3.9\times10^{-12}\ M \end{align*} ]

Answer:

(3.9\times10^{-12})