pH - AP Chemistry

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.

For a solution to be acidic, it must have a pH between 1 and 6.99, since 7 is neutral; pH is –log(hydrogen ion concentration), the range of possible values is between 10–1 and 10–6.999.

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.

For a solution to be acidic, it must have a pH between 1 and 6.99, since 7 is neutral; pH is –log(hydrogen ion concentration), the range of possible values is between 10–1 and 10–6.999.

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.

For a solution to be acidic, it must have a pH between 1 and 6.99, since 7 is neutral; pH is –log(hydrogen ion concentration), the range of possible values is between 10–1 and 10–6.999.

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.

pOH would be 4 (use –log \[OH–\]) and pH would be 14–pOH = 14 – 4 = 10

\[OH-\] = 4 X 10-6

pOH = 5.4 — use –log \[OH–\] to find pOH

pH = 14– pOH = 8.6

The pH of a solution is determined by taking the negative log of the concentration of hydrogen ions in solution. HCl is strong acid so it completely dissociates in solution. So adding .0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log\[.0001\] =4, so the pH of the solution =4.

\[H+\] = 3.2 X 10-7 mol

pH = -log \[H+\]

= 7 - log 3.2

= 7 -.505

=6.495

pOH = 14 - pH

= 7.5

This question asks for the solution with the greatest number H+ ions, we can also approach this problem as if we are looking for the solution with the lowest pH. HCl is strong acid, therefore it will dissociate completely in solution. So the solution containing 0.010 mL of HCL contains 0.010 mL of H+ ions in solution, in addition to the H+ ions that are already in solution due to the auto-ionization of water. The only other solution that could have a pH less than 7 would be the one with 0.010 mL of CH3OH in excess water, becasue CH3OH is very slightly acidic. But since it is compared with an equal volume of HCl which is strong acid, it can be said that the most H+ ions will be found in the solution containing a small amount of strong acid, HCl.

Calcium carbonate is a base since it's the salt of a strong base (NaOH) and a weak acid (carbonic acid). The only basic pH on the list is 8, making it the correct answer.

The –log of the pKa will give you the Ka, so take the –log (6.4), which gives you approximately 4 * 10–7. The Ka expression is set up with products over reactants (hydrogen carbonate ion * hydrogen ion/carbonic acid). The undissociated carbonic acid is 0.001M, and you should use the variable 'x' to account for how much it dissociates and how many of the ions are produced. Ka = 4 * 10–7 = x2/0.001 ends up being your Ka expression, if you assume x is negligible compared to the original concentration of 0.001. Solving for x, you get 2 * 10–5. This is the hydrogen ion concentration. pH = –log (hydrogen ion concentration), so pH = –long(2 * 10–5), which is approximately 4.7.

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

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 = _x_2/.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.

Consider the reaction of and :

Now we will calculate the moles of in the solution prior to adding base.

We will then calculate the amount of moles of that react with the base.

We will then calculate the remaining moles of :

We will then calculate the new concentration of sulfuric acid:

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

First, we will calculate as follows:

Now we will calculate the pOH:

In solution , so .