A solution is prepared by adding 0.10 Mol of sodium fluoride, NaF, to 1.00 L of

Question

A solution is prepared by adding 0.10 Mol of sodium fluoride, NaF, to 1.00 L of water. Which statement about the solution is correct? (See #17 in photo for choices)

16. What isthe value of Kb for the cyanide anion, CN'? Ka(HCN) = 6.2 × 10-10 1.6 x 104 1.6 × 10-5 c) 3.8 x 104 d) 3.8 x 10 5 e) 6.2 × 104 17. A solution is prepared by adding 0.10 mol of sodium fluoride, NaF, to 1.00 L of water. Whi statement about the solution is correct? a) The solution is basic. b) The solution is neutral. c) The solution is acidic. d) The concentrations of fluoride ions and sodium ions will be identical. e) The concentration of fluoride ions will be greater than the concentration of sodium ions. 18. Formic acid, which is a cormponent of insect venom, has a K, hich is a component of insect venom, has a Ka 1.8 x 10. What is the [H,O 1.8 x 10". What is the [H ,0 ution that is initially 0.10 M formic acid, HCOOH? a.210M b) 8.4 × 10° M

Explanation / Answer

Question No. 17 :

Given : mol of NaF = 0.10 mol

Volume of water= 1.00 L

We know NaF is salt and it dissociates completely.

[Na+] = [ F-] = 0.10 mol / 1.00 L = 0.10 M

We know F- again reacts with water and forms OH- in the solution.

Lets show the ICE chart

            F-(aq)   + H2O (l)         ----- > HF (aq) + OH- (aq)

I           0.10 M                                                 0                      0

C          -x                                             +x                    +x

E          (0.10 –x)                                  x                      x

We use kb value of F-

Kb = 1.0E-14 / ka

Ka for HF = 7.2E-4

Kb = 1.0E-14 / 7.2E-4

= 1.39E-11

We use this kb to get x

1.39E-11 = x2/ ( 0.1 –x)

We neglect x in the denominator since value of kb is very small

1.39E-11 = x2/ 0.1

x = 1.18 E-16

x = [OH-]= 1.18E-6

pOH = -log [OH-]

= -log ( 1.18E-6)

= 5.92

Now pH = 14-pOH = 14 -5.92 = 8.07

pH is basic since it is more than 7

So answer for this question is option a)

The solution is basic

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