Consider the reaction of \(\displaystyle NaOH\) and \(\displaystyle H_2SO_4\):

\(\displaystyle 2NaOH+H_2SO_4\rightarrow Na_2SO_4\hspace{1 mm}+2H_2O\)

Now we will calculate the moles of \(\displaystyle H_2SO_4\) in the solution prior to adding base.

\(\displaystyle 500\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{16\hspace{1 mm}moles\hspace{1 mm}H_2SO_4}{1\hspace{1 mm}L}=8.00\hspace{1 mm}moles\hspace{1 mm}H_2SO_4\)

We will then calculate the amount of moles of \(\displaystyle H_2SO_4\) that react with the base.

\(\displaystyle 625\hspace{1 mm}g\hspace{1 mm}NaOH\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaOH}{40\hspace{1 mm}g\hspace{1 mm}NaOH}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}H_2SO_4}{2\hspace{1 mm}moles\hspace{1 mm}NaOH}=7.81\hspace{1 mm}moles\hspace{1 mm}H_2SO_4\)

We will then calculate the remaining moles of \(\displaystyle H_2SO_4\):

\(\displaystyle 8.00\hspace{1 mm}moles\hspace{1 mm}H_2SO_4 -\hspace{1 mm}7.81\hspace{1 mm}moles\hspace{1 mm}H_2SO_4=0.19\hspace{1 mm}moles\hspace{1 mm}H_2SO_4\)

We will then calculate the new concentration of sulfuric acid:

\(\displaystyle \frac{0.19\hspace{1 mm}moles\hspace{1 mm}H_2SO_4}{0.5\hspace{1 mm}L}=0.38\hspace{1 mm}M\)

Sulfuric acid is a diprotic acid, so the hydrogen ion concentration is 0.76 M.

\(\displaystyle pH=-log[H^{+}]=-log(0.76)=0.12\)

The expression for Ksp is Ksp = [Mg₂₊][OH–]².

Thus, [OH–] = √(1.2 * 10^–11)/(1.2 * 10^–5)

[OH–] = 1 * 10^–3

Thus, pH = –log(1 X 10^–3) = 3 

pH = 14 – 3 = 11

Since the solution has a pH of 2.44, we can find the concentration of protons in the solution using the pH equation.

\(\displaystyle pH=-\log[H^+]\rightarrow [H^+]=10^{-pH}\)

\(\displaystyle \small \small \small [H^{+}] = 10^{-2.44} = 3.6*10^{-3}M\)

All of these acids are monoprotic acids, so the equilibrium expression for each acid follows the formula for dissociation.

\(\displaystyle HA\rightarrow H^++A^-\)

\(\displaystyle \small K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\)

The concentration for protons will be equal to the concetration of the conjugate base, since they are both in a one-to-one ratio with the acid molcule.

\(\displaystyle [H^+]=[A^-]=3.6*10^{-3}M\)

Also, we can subtract the concentration of protons from the initial concentration of acid in order to find how much acid will remain at equilibrium.

\(\displaystyle [HA]_{initial}-3.6*10^{-3}M=[HA]_{final}\)

Using the terms for each concentration, we can try to solve the equilibrium expression.

\(\displaystyle \small \small K_{a} = \frac{[3.6*10^{-3}][3.6*10^{-3}]}{[1-3.6*10^{-3}]}\)

\(\displaystyle K_a=\frac{1.3*10^{-5}}{0.9964}=1.3*10^{-5}\)

Since the acid dissociation constant is equal to \(\displaystyle \small 1.3*10^{-5}\), we can look at the list and determine that propanoic acid was used to make the acidic solution.

Since every calcium hydroxide molecule will dissociate and form two hydroxide ions, the concentration of the hydroxide ions will be twice as much as the initial concentration of the base.

\(\displaystyle 0.2M\ Ca(OH)_2*\frac{2mol\ OH^-}{1mol\ Ca(OH)_2}=0.4M\ OH^-\)

This means that the concentration of hydroxide ions in the solution will be 0.4M.We can use this concentration to solve for the pOH fo the solution.

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

\(\displaystyle pOH=-\log(0.4)=0.4\)

Since we are looking for the pH of the solution, we simply subtract the pOH form 14.

\(\displaystyle pH+pOH=14\rightarrow pH=14-pOH\)

\(\displaystyle pH=14-0.4=13.6\)

This results in a pH of 13.6.

Basic solutions have a pH level greater than 7, which is the neutral pH. Acidic solutions have a pH level lower than 7.

Basic solutions have a pH level greater than 7 and acidic solutions have a pH level lower than 7. A pH of 7 corresponds to a neutral solution.

Acidic solutions have a pH level lower than 7, which is the neutral pH. Basic solutions have a pH level greater than 7.

Potassium sulfide is soluble in water yielding the specie \(\displaystyle S^{-2}\) which undergoes a two steps basic hydrolysis:

\(\displaystyle S^{-2} + H_{2}O \rightarrow HS ^{-1} + OH^{-}\)

\(\displaystyle HS^{-} + H_{2}O \rightarrow H_{2}S + OH^{-}\)

The respective hydrolysis constants are:

\(\displaystyle K_{H1} = \frac{K_{w}}{K_{a2}}= \frac{10^{-14}}{10^{-19}}= 10^{5}\)

\(\displaystyle K_{H2} = \frac{K_{w}}{K_{a1}}= \frac{10^{-14}}{9\times 10^{-8}}= 10^{7}\)

Analyzing the rounded values of both hydrolysis constants, we can see that the value of \(\displaystyle K_{H1}\) is very large then we can assume the first reaction occurs to completion and controls the concentration of \(\displaystyle OH^{-}\) in the solution which will be \(\displaystyle 0.01 M\). Hence, the pH will be:

 \(\displaystyle pH = 14-pOH = 14 - \left ( -\log_{10}0.01 \right ) = 14 - 2 = 12\)

This question is presenting us with a reversible chemical reaction of a weak acid. It provides us with the pKa of the acid, and asks us to determine where on the pH spectrum will the acidic form of acetic acid be equal to its basic form.

There is a very important concept that this question highlights, which is that a compound will always have equal amounts of its acidic and basic forms at a pH that is equal to its pKa. This can be shown mathematically by using the Henderson-Hasselbalch equation.

\(\displaystyle pH=pKa+log\left ( \frac{Base}{Acid} \right )\)

As the above equation shows, when the acidic and basic forms are equal, their ratio is equal to \(\displaystyle 1\). And the logarithm of \(\displaystyle 1\) is equal to \(\displaystyle 0\). Thus, what we're left with is:

\(\displaystyle pH=pKa\)

Dipole moments are the result of the unequal distribution of electrons in a bond or in a molecule.  Electrons will move towards the more electronegative atoms, leading to a unequal distribution of charge refered to as a dipole moment.  The answer here if HF because the flourine will strongly overpower the H atom's ability to attract electrons and the molecule will have a very large seperation of charge with the F end being partially egative and the H end being partially positive. This dipole will be stronger than that of HCl becasue F is more electronegative than HCl.