Chemistry: The Central Science (13th Edition)

The molarity of the solution is equal to 4.5 M $NaCH_3COO$
1. $Na^+$ is not an electrolyte, but $CH_3COO^-$ is. 2. Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_b = 5.6\times 10^{- 10}$ 3. Find the $[OH^-]$ value: $pH + pOH = 14$ $pOH = 14 - pH = 14 - 9.70 = 4.30$ $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{-4.30}$ $[OH^-] = 5.0 \times 10^{-5}$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: $CH_3COO^-(aq) + H_2O(l) \lt -- \gt CH_3COOH(aq) + OH^-(aq)$ \begin{matrix} & [CH_3COO^-] & [OH^-] & [CH_3COOH] \\ Initial & [Initial \space CH_3COO^-] & 0 & 0 \\ Change & -x & +x & +x\\ Equil & [Initial \space CH_3COO^-] -x & x & x \end{matrix} 5. Now, use the Kb and x values and equation to find the initial concentration value. $K_b = \frac{[OH^-][CH_3COOH]}{ [Initial \space CH_3COO^-] - x}$ $5.6\times 10^{- 10}= \frac{[x^2]}{ [Initial \space CH_3COO^-] - x}$ $5.6\times 10^{- 10}= \frac{( 5.0 \times 10^{-5})^2}{[Initial \space CH_3COO^-] - 5.0 \times 10^{-5}}$ $[Initial \space CH_3COO^-] - 5.0 \times 10^{-5}= \frac{( 5.0 \times 10^{-5})^2}{ 5.6\times 10^{- 10}}$ $[Initial \space CH_3COO^-] = \frac{( 5.0 \times 10^{-5})^2}{ 5.6\times 10^{- 10}} + 5.0 \times 10^{-5}$ $[Initial \space CH_3COO^-] = 4.5$ 6. $[NaCH_3COO] = [CH_3COO^-]$. Thus: $[NaCH_3COO] = 4.5 \space M$