Question

question 6 (5 points)
what is the ph of a 0.340 mol/l solution of sodium fluoride?
(k_a hf = 7.2 x 10^{-4})
ph=3.0
ph=8.3
ph=8.2
ph=5.6

Explanation:

Step1: Write the hydrolysis reaction

Sodium fluoride ($NaF$) is a salt of a weak - acid ($HF$) and a strong - base ($NaOH$). The fluoride ion ($F^-$) hydrolyzes in water: $F^-(aq)+H_2O(l)
ightleftharpoons HF(aq) + OH^-(aq)$. The base - dissociation constant $K_b$ for $F^-$ can be calculated from the acid - dissociation constant $K_a$ of $HF$ using the relationship $K_w = K_a\times K_b$, where $K_w=1.0\times 10^{-14}$.

Step2: Calculate $K_b$

Given $K_a = 7.2\times 10^{-4}$, then $K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-4}}\approx1.39\times 10^{-11}$.

Step3: Set up the equilibrium expression

Let $x$ be the concentration of $OH^-$ and $HF$ at equilibrium. The initial concentration of $F^-$ is $c = 0.340\ mol/L$. At equilibrium, $[F^-]=0.340 - x$, $[HF]=x$, and $[OH^-]=x$. Since $K_b$ is very small, we can assume that $0.340 - x\approx0.340$. The equilibrium expression for the hydrolysis of $F^-$ is $K_b=\frac{[HF][OH^-]}{[F^-]}$. Substituting the values, we get $1.39\times 10^{-11}=\frac{x\cdot x}{0.340}$.

Step4: Solve for $x$ (concentration of $OH^-$)

$x^2=K_b\times[F^-]=1.39\times 10^{-11}\times0.340$. So, $x = [OH^-]=\sqrt{1.39\times 10^{-11}\times0.340}\approx2.16\times 10^{-6}\ mol/L$.

Step5: Calculate $pOH$ and $pH$

$pOH=-\log[OH^-]=-\log(2.16\times 10^{-6})\approx5.67$. Since $pH + pOH=14$, then $pH = 14 - pOH=14 - 5.67 = 8.33\approx8.3$.

Answer:

$pH = 8.3$