## Chemistry and Chemical Reactivity (9th Edition)

$pH = 3.85$
1. Calculate the molar mass $(NaCH3CO2)$: 22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 2 = 82.04g/mol 2. Calculate the number of moles $(NaCH3CO2)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 1.56}{ 82.04}$ $n(moles) = 0.0190$ 3. Find the concentration in mol/L $(NaCH3CO2)$: $0.019$ mol in 1L: $0.019 M (NaCH3CO2)$ 4. Drawing the ICE table, we get these concentrations at the equilibrium: $CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3C{O_2}^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[CH_3CO_2H] = 0.15 M - x$ $[CH_3C{O_2}^-] = 0.019M + x$ $[H_3O^+] = 0 + x$ 5. Calculate 'x' using the $K_a$ expression. $1.8\times 10^{- 5} = \frac{[CH_3C{O_2}^-][H_3O^+]}{[CH_3CO_2H]}$ $1.8\times 10^{- 5} = \frac{( 0.019 + x )* x}{ 0.15 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.019 * x}{ 0.15}$ $1.8\times 10^{- 5} = 0.13x$ $\frac{ 1.8\times 10^{- 5}}{ 0.13} = x$ $x = 1.4\times 10^{- 4}$ Percent dissociation: $\frac{ 1.4\times 10^{- 4}}{ 0.15} \times 100\% = 0.095\%$ x = $[H_3O^+]$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.4 \times 10^{- 4})$ $pH = 3.85$