The Ka = [H+][F–]/[HF]. The beginning concentration of HF is given as 0.05, and we can use x as the variable that accounts for how much of the HF is lost, along with how much of the H+ and F– are formed. 

Thus, Ka = x²/.05 if you use the approximation that x is negligible compared to the starting concentration of HF. 

Solving for x, x = 1 * 10^–3. This is the H+ ion concentration.

pH = –log(H+) = –log(1.0 * 10^–3)) = 3

NaOH is a base, so that won't produce an acidic solution. Of the remaining acids, HCl and HI are strong acids, and HF is weak. HI is at a higher molarity, so it will produce the most acidic solution. 

First, we will calculate \(\displaystyle [KOH]\) as follows:

\(\displaystyle \frac{4.2\hspace{1 mm}g\hspace{1 mm}KOH}{500\hspace{1 mm}mL}\times\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}KOH}{56.1\hspace{1 mm}g\hspace{1 mm}KOH}=1.497\times10^{-1}\hspace{1 mm}M\)

Now we will calculate the pOH:

\(\displaystyle pOH=-log[OH^-]\)

In solution \(\displaystyle KOH_{(s)}\rightarrow K^{+}_{(aq)}+OH^{-}_{(aq)}\), so \(\displaystyle [KOH]=[OH^-]\).

\(\displaystyle pOH=-log[1.497\times 10^{-1}]=8.248\times 10^{-1}\)

\(\displaystyle pH=14-pOH=14-0.8248=13.18\)

The pH of the solution is 10.2, and we know:

\(\displaystyle pH\hspace{1 mm}+\hspace{1 mm}pOH=14\)

\(\displaystyle pOH=14-pH=14-10.2=3.8\)

\(\displaystyle pOH=-log[OH^-]\)

\(\displaystyle 3.8=-log[OH^-]\)

\(\displaystyle -3.8=log[OH^-]\)

\(\displaystyle [OH^-]=10^{-3.8}=1.58\times 10^{-4}\)

And since \(\displaystyle NaOH_{(s)}\rightarrow Na^+_{(aq)}+OH^-_{(aq)}\), \(\displaystyle [NaOH]=1.58\times 10^{-4}\).

\(\displaystyle 627\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{1.58\times 10^{-4}\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=4.03\times 10^{-3}\hspace{1 mm}g\hspace{1 mm}NaOH\)

First, we will determine the amount of hydronium ions in the first solution. This allows us to measure the moles of hydronium ion that are transferred to water.

\(\displaystyle pH=-log[H^+]\)

\(\displaystyle 3.00=-log[H^+]\)

\(\displaystyle [H+]=10^{-3}=0.00100\hspace{1 mm}M\)

\(\displaystyle 10\hspace{1 mm}mL\times\frac{1\hspace{ 1 mm}L}{1000\hspace{ 1 mm}mL}\times\frac{0.001\hspace{ 1 mm}mol}{1\hspace{ 1 mm}L}=1.00\times10^{-5}\hspace{ 1 mm}mol\)

Now we can calculate the hydronium ion concentration in the new solution, using the amount transferred and the volume of water, and can solve for pH.

\(\displaystyle \frac{1.0\times10^{-5}mol}{60\hspace{ 1 mm}mL}\times\frac{1000\hspace{ 1 mm}mL}{1\hspace{ 1 mm}L}=1.67\times 10^{-4}\hspace{ 1 mm}M\)

\(\displaystyle pH=-log[1.67\times 10^{-4}]=3.78\)

HCl is a strong acid, while NaCl is a salt. Mixing these two will result in a pH < 1.

As for the other choices, mixing 2.0M HCl and 2.0M KOH will result in the formation of KCl and H₂O, resulting in a neutral pH. Acetic acid (pKa= 4.76) is a weak acid, so it will have a pH less than 7 but greater than 1. Finally, NH₃ and NH₄⁺ is a buffer system between a weak base and weak acid, and if equal amounts are present, the pH will be the same as the pKa of NH₄⁺, which is around 9.

The pH of a solution is given by the equation \(\displaystyle pH=-log[H^+]\).

1) We can assume that hydrobromic acid, as a strong acid, will dissociate completely. As a result, there will be 1mol of protons in the initial solution. Molarity is a method of measuring the concentration of an agent in a solution and is defined by the equation below.

\(\displaystyle M=\frac{mol\ solute}{L\ solution}\)

Since we have 1mol of protons and 1L of solution, the molarity of protons is simply 1M in the initial solution (1mol/1L). We can plug this into the pH equation and solve.

\(\displaystyle pH=-log[1]=0\)

2) After the dilution with water, the concentration of protons will change. Since the volume of the solution has doubled, the concentration of protons has been halved.

\(\displaystyle M=\frac{1mol}{2L}=0.5M\)

\(\displaystyle pH=-log[H^+]=-log[0.5]=0.3\)

* Key points to remember!

1) The lower the pH, the more acidic the solution. This is why the diluted solution was 0.3 and the original, more acidic solution was 0.

2) pH is a logarithmic scale. As a result, a pH of 2 is ten times more acidic than a pH of 3. This is why halving the concentration DID NOT halve the pH.

We first need to calculate the moles of NaOH in 10mL of a 0.56M solution.

\(\displaystyle 10\hspace{1 mm}mL\cdot\frac{0.56\hspace{1 mm}mol\hspace{1 mm}NaOH}{1000\hspace{1 mm}mL}=5.6\cdot 10^{-3}\hspace{1 mm}mol\hspace{1 mm} NaOH\)

Now, in the new volume, the concentration is equal to this same number of moles divided by the new volume. The new volume is the sum of the two solutes together, so the sum of 10mL of solution and 100mL of water.

\(\displaystyle \frac{5.6\cdot 10^{-3}\hspace{1 mm}mol\hspace{1 mm}NaOH}{110\hspace{1 mm}mL}\cdot\frac{1000\hspace{1 mm}mL}{1\hspace{1 mm}L}= 5.09\cdot 10^{-2}\hspace{1 mm}M\)

Now we need to calculate pH. Since NaOH will dissociate one hydroxide ion for every molecule of NaOH in solution, the concentration of hydroxide is equal to the concentration of NaOH.

\(\displaystyle [NaOH]=[OH^-]=5.09\cdot 10^{-2}M\)

\(\displaystyle pOH=-log[OH^-]=-log[5.09\cdot 10^{-2}]=1.29\)

Knowing the pH allows us to solve for the pH.

\(\displaystyle pH+pOH=14\)

\(\displaystyle pH=14-pOH=14-1.29=12.71\)

[H⁺] = 3.2 X 10^-7 mol

pH = -log [H⁺]

= 7 - log 3.2

= 7 -.505

=6.495

pOH = 14 - pH

= 7.5

For a solution to be basic, it must have a pH between 8 and 14, since 7 is neutral; pH is –log(hydrogen ion concentration); the range of possible values is between 10^–13 and 10^–7.001.