Calculate the pH of a 0.200-M solution of C5 H5 NHF. Hint: C5 H5 NHF is a salt composed of C5 H5

Calculate the pH of a 0.200-M solution of C5 H5 NHF. Hint:...

Calculate the pH of a $0.200-M$ solution of $\mathrm{C}_5 \mathrm{H}_5 \mathrm{NHF}$. Hint: $\mathrm{C}_5 \mathrm{H}_5 \mathrm{NHF}$ is a salt composed of $\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+}$and $\mathrm{F}^{-}$ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$ \begin{array}{r} \mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \\ \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3} \end{array} $$

Calculate the pH of a $0.200-M$ solution of $\mathrm{C}_5 \mathrm{H}_5 \mathrm{NHF}$. Hint: $\mathrm{C}_5 \mathrm{H}_5 \mathrm{NHF}$ is a salt composed of $\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+}$and $\mathrm{F}^{-}$ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is
$$
\begin{array}{r}
\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \\
\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}
\end{array}
$$

Key Concepts

pH Calculation from Equilibrium

Calculating the pH of a solution involving acid-base equilibria requires setting up an equilibrium expression that accounts for the concentrations of all species involved. By applying the equilibrium constant and considering the initial concentrations and changes during the reaction, one can solve for the hydrogen ion concentration ([H+]) and thus the pH. This process often involves simplifying assumptions or approximations to facilitate the calculation.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are formed when an acid loses a proton to become a base, or when a base gains a proton to become an acid. This pairing is a fundamental concept in acid-base chemistry, as the relative strengths of the conjugate pairs dictate the direction of equilibrium in proton transfer reactions. Analyzing these pairs helps in determining which species will predominate in solution and how they will influence the pH.

Equilibrium Constant

The equilibrium constant (K) quantifies the position of equilibrium in a reversible chemical reaction. For acid-base reactions, it relates the concentrations of the reactants and products at equilibrium, providing insight into the favorability of the forward or reverse reaction. A smaller K value indicates that the reaction favors the reactants, which is important when assessing the extent of ionization or proton transfer in solution.

Acid-Base Equilibrium

Acid-base equilibrium involves reversible reactions where acids donate protons and bases accept them. In aqueous solutions, the extent of proton transfer between acids and bases determines the concentrations of various species, which in turn affect the pH. Understanding these equilibria is essential for predicting the behavior of compounds in solution.

Salt Hydrolysis

Salt hydrolysis occurs when the ions from a dissolved salt react with water to form either an acidic or a basic solution. This process involves one ion acting as an acid (donating a proton) or a base (accepting a proton), thereby shifting the pH away from neutrality. Recognizing and analyzing salt hydrolysis reactions is key for calculating the pH of solutions containing salts derived from weak acids or bases.

Transcript

00:04 Let's calculate the ph of a .2 molar solution of c5h5nhf, which will dissociate to c5h5nh plus and f minus and 0 .2 molar, and 0 .2 molar and 0 .2 molar and 0 .2 molar right across.

00:36 So now set up our equilibrium, which we're given in the question.

00:50 And this would yield c5h5n and hf.

01:02 The initial set up a nice table here, 0 .2 and 0 .0 minus x plus x plus x.

01:14 This would yield 0 .2 minus x, 0 .2 minus x, x and x.

01:25 The equilibrium constant here is defined as c5h5n, hf, divided by c5h5nh plus and f minus.

01:48 Now we have to solve for the equilibrium constant here.

01:54 The equilibrium constant will be equal to the equilibrium, the k -a for c -5 -h -5n -h plus times the, it's going to be the kb, so 1 over the k -a for h -f.

02:16 So let's plug these values in here.

02:18 We'll get 5 .9 times 10 to the negative 6 times 1 over 7 .2 times 10 to the negative 4 and our ka will work out to, or our equilibrium not our ka, but our equilibrium constant works out to 8 .2 times 10 to the negative 3.

02:39 So let's use this on our ice table, substitute everything in.

02:42 8 .2 times 10 to the negative 3 is equal to x .x .6 .0 .2 .2.

02:50 2 minus x, 0 .2 minus x.

02:55 Let's rewrite this.

02:57 8 .2 times 10 to the negative 3 is equal to x squared.

03:03 0 .2 minus x.

03:06 Take the square root of both sides and we will find that we get 0 .091 is equal to x over 0 .2 minus x...