Be sure to answer all parts. A 35.00−mL solution of 0.2500 M HF is titrated with a standardized 0.1318 M solution of NaOH at 25°C. (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? mL (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?

A) 0.60 is the pH of the HF solution before titrant is added.

b) 66.38 ml is the amount of titrant required to reach equivalence point.

c)9.1 is the pH at 0.50 mL before the equivalence point.

d)9.1 is the pH at the equivalence point.

e)  8.9 is the pH at 0.50 mL after the equivalence point.

Explanation:

Data given:

a) volume of HF = 35 ml

concentration of HF = 0.25 M

pka of HF = 3.14

pH = ?

HF is an strong acid so it ionizes completely thus hydronium ion concentration is 0.25 M

pH = - log [H3O+]

pH = - log [ 0.25]

      = - [- 0.60]

pH = 0.60

b) To calculate volume of titrant, formula used is

MacidVacid = MbaseVbase

volume of acid = 35ml

molarity of acid = 0.25

molarity of titrant base = 0.1318

volume of base =?

putting values in the equation

35 x 0.25 = Vbase x 0.1318

V base =

            = 66.38 ml

c) pH at 0.50 ml before equivalence point

molarity of NaF is 0.25 before 0.50 ml equivalence point.

molarity of NaF =

                           =

               [NaF]   = 0.086

pH = 7+.(pKa -log [NaF])

pH = 9.1

d) pH at equivalence point

molarity of NaF =

                          =

        molarity    = 0.086

pH = 7+.(pKa -log [NaF])

     = 7 + O.5  (3.14 + 1.06)

      = 7 + 2.1

pH   = 9.1

e) pH at 0.50 ml after equivalence point is

molarity of NaF is calculated by:

molarity of NaF =

                          = 0.172

pH = 7+.(pKa -log [NaF])

   pH    = 8.9

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: