What is the effect on equilibrium when sodium format is added to a solution of f

Question

What is the effect on equilibrium when sodium format is added to a solution of formic acid? no change in the equilibrium equilibrium shifts to the right more information is needed to answer the question equilibrium shifts to the left Which of the following aqueous mixtures would be a buffer system? HCl, NaCl HNO_3, NaNO_3 H_3PO_4, NaH_2PO_4 H_2SO_4, CH_3COOH NH_3, NaOH A 35.0 mL sample of 0.20 M LiOH is titrated with 0.25 M HCl. What is the pH of the solution after 23.0 mL of HCl have been added to the base? For PbCl_2 (K_sp = 2.4 times 10^-4), will a precipitate of PbCl_2 form when 0.10 L of 3.0 times 10^-2 M Pb(NO_3)_2 is added to 400 mL of 9.0 times 10^-2 M NaCl? SHOW WORK TO RECEIVE CREDIT!! Yes. because Q > K_sp. No. because Q

Explanation / Answer

1. Formic acid is dissociated as this:

HCOOH <--> H+ + COOH-

As you add Sodium formate, you are adding basically Formate (COOH-) to your solution; so to equilibrate your reaction it tends to go to THE LEFT (E)

2. A good buffer system has to be weak acid/base with it`s conjugate base/acid, only b and c are conjugated base/acid, and only C is a weak acid, HNO3 is a strong acid, so its not an answer

3. First we have to find the moles (quantity) of every reactant

LiOH= 35.0mL Volume=; Concentration=0.20 mol/L; Moles= Concentration x Volume in L= 0.20x35x10-3=0.007 moles of LiOH

   HCl= 23. mL Volume=; Concentration=0.25 mol/L; Moles= Concentration x Volume in L= 0.25x23x10-3= 0.00575 moles of HCl

Let`s remember that both reactants will dissociate at 100%. Also as you can see above there are more OH than H, so, all the H will be neutralized, there will be OH left:

0.007- 0.00575= 0.00125 moles of OH left

Now concentration of OH= Moles/ TOTAL Volume in L (35mL+23mL)= 0.00125/ 0.058= 0.02155 M of OH

pOH= -log(OH)

pOH= 1.66

pH= 14-pOH

pH= 12.33

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