2500 ml of an solution of naf of unknown concentration was...

25.00 mL of an solution of NaF of unknown concentration was titrated with 0.222 M HCl. The titration required 12.34 mL of HCl to reach the equivalence point.Ka of HF = 7.2•10-4a. Calculate the initial concentration of NaF.b. Calculate the initial pH of the NaF solution.c. What volume of HCl is required to have pH = pKa?d. Calculate the pH of the solution after 10.12 mL of HCl were added.e. Calculate the pH of the solution after 12.34 mL of HCl were added.f. Calculate the pH of the solution after 15.67 mL of HCl were added

Transcript

00:01 This is a strong acid, weak base titration.

00:08 And so we have a kb value for the weak base that we are titrating with the strong acid.

00:16 So because it is a weak base that we start with in solution, before adding any strong acid, we are going to have a basic ph.

00:26 The basic ph can be calculated from the kb equation, rearranged to solve for hydroxide, where the hydroxide concentration is going to be equal to the kb value multiplied by the concentration of the base, and both of those were given to us.

00:46 Taking the square root, we get 4 .69 times 10 to the negative 5, so the ph is going to be the negative log of the hydronium concentration, which we can get from the hydroxide concentration, by dividing the hydroxide concentration into kw, and then we take the negative log of that to get our ph at 9 .67.

01:07 And as you can see, there's a basic ph above 7.

01:12 After adding 25 milliliters, we are still pre -equivalence.

01:17 We have not yet reached the equivalence point.

01:19 We recognize that because we have 100 milliliters of .2 molar weak base, and we are adding a strong acid with half the concentration of the weak base.

01:29 So we are going to need double the volume of the strong acid.

01:33 Acid to neutralize 100 milliliters of the weak base or 200 milliliters.

01:39 So 25 milliliters is definitely pre -equivalence.

01:43 Pre -equivalence, we have a buffer solution, and for a buffer solution, we can use the henderson -hasselbalch equation to solve for ph, where ph is going to be equal to pca, which is the negative log of the k -a value, which we can get by taking the kb value, which was given to us, and dividing it into kw, plus the lh, which is going to the negative.

02:03 Log of the moles of base over the moles of acid.

02:07 The moles of base that we have left after adding 25 milliliters is going to be equal to the moles of base we start with, which will be its volume multiplied by its molarity, minus the moles of base that reacted, which will be equal to the volume multiplied by the molarity of the strong acid added.

02:25 For every one mole of strong acid added, we consume one mole of weak base, and in the process, make one mole of weak acid.

02:34 Therefore, the moles of weak acid formed will simply be equal to the moles of strong acid added.

02:39 And we get a ph of 6 .89.

02:42 Now at 70 milliliters, we're still pre -equivalents...

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