# chemistry

posted by .

if the pH of a half-neutralized acid solution is 5.4, how would you find the [H+] of the the solution. There is 1.0 g of L-ascorbic acid which was dissolved in 100 ml of water. That solution was split in two and 50 ml of the solution was titrated with 0.2 M NaOH (13 ml NaOH). then the other half of the solution was added to get the pH of 5.4. With this information how can I get the concentration of H+?

• chemistry -

If the pH = 5.4 then
5.4 = -log(H^+)
-5.4 = log(H^+)
(H^+) = 3.98 x 10^-6
I didn't work through all the numbers to see if they match up but it doesn't matter what they are. If the pH is 5.4, then (H^+) is defined by that and nothing else.

• chemistry -

oh ok.. thanks so much... So when you have the pH of a substance the [H^+] is always equal to 10^-pH?

• chemistry -

is this then equal to the Ka of the acid solution?

• chemistry -

Yes to the first question. (H^+) = 10^-pH.
At the half way point, which is what you had, then Ka = (H^+ or pKa = pH but that is only true at the half way point. You can see for a weak acid, such as HA
HA ==> H^+ + A^- and
Ka(H^+)(A^-)/(HA). Solving for (H^+) we get
(H^+) = Ka(HA)/(A^-). SO, when exactly half the HA has been neutralized, then there has been formed an equal amount of (A^-) so (HA)=(A^-) and (H^+) = Ka. You can take the -log of each side to obtain
-log(H^+) = -log Ka AND
pH = pKa. But remember, for this second part, this is true ONLY at the half way point to neutralization of the acid (Or, of course, when you make a buffer solution containing equal amounts of acid and salt).

## Similar Questions

1. ### chemistry

Ascorbic acid is a weak organic acid also known as vitamin C. A student prepares a 0.20 mol/L aqueous solution of ascorbic acid and measures its pH as 2.40. Based on this evidence, what is the Ka of ascorbic acid?
2. ### college-chemistry

2. A 0.1967 g sample of pure ascorbic acid is weighed out and dissolved in 50.0 mL of distilled water in an Erlenmeyer flask. One week later, when about one third of the water in the flask has evaporated, this solution is titrated …
3. ### chemisrty

Q1) A solution of ascorbic acid (C6H8O6, Formula mass = 176 g/mol) is made up by dissolving 80.5 g of ascorbic acid in 210 g of water, and has a density of 1.22 g/mL at 55 oC. Calculate the following: a) the mass percentage of ascorbic …
4. ### urgent i need help within an hour!

Ascorbic acid (vitamin C, C6H8O6) is a water-soluble vitamin. A solution containing 83.3 g of ascorbic acid dissolved in 210. g of water has a density of 1.23 g/mL at 55°C Calculate the molarity of ascorbic acid in this solution. …
5. ### College Chemistry

You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point. The pH of the other half of the original …
6. ### chemistry

You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point. The pH of the other half of the original …
7. ### chemistry

You weigh a sample of a monoprotic unknown acid and dissolve it in 50.00 mL of distilled water. Exactly half of this solution is titrated with Sodium Hydroxide to the phenolphthalein end point. The pH of the other half of the original …
8. ### chemistry

How would you prepare a solution of about 8mg/100ml of ascorbic acid in water (25ml)?
9. ### chemisrty

A solid acid is dissolved in enough water to make 200 ml of a solution. 40.0ml of the solution is titrated to a phenolphthalein en point with an NaOH solution. The neutralized solution and the remainder of the acid solution are then …
10. ### chemistry

How would you do this question: A student dissolved 5.0 g of vitamin C in 250 mL of water. The molar mass of ascorbic acid is 176 g/mol, and its Ka is 8.0 × 10−5. Calculate the pH of the solution. Note: Abbreviate the formula …

More Similar Questions