Question

Name Constants: R0.0821 Latm mold atm, 0°C = 273.15 K, K 1x10 at 25°C b 52610 , C3 .0 x 10 , h = 6.626 x 10 3.0 x 10 m/s, 760 mm Hg = 1 Short Answer Questions: Show ALL WORK to receive partial credit. NO WORK, NOCH ceive partial credit. NO WORK, NO CREDITI 1. Determine the pH for a 0.267 M NaCH,H,O, solution at 25 °C, the K, for HOH (10 points) CHH,O, solution at 25 °C, the K, for HO,H.CH, is 1.38 x 10

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Answer 1

1. Ans :-

Given,

Initial concentration of Base (say A^-) = 0.267 M

ICE table is :

...................................A^- ...................+...................H₂O <-------------------> HA ................+.....................OH^-

Initial .........................0.267 M.....................................................................0.0 M....................................0.0 M

Change.........................-y...............................................................................+y........................................+y

Equilibrium ................(0.267-y) M....................................................................y M.......................................y M

Where,

y = Amount dissociated per mole

Expression of base dissociation constant (K_b) is :

K_b = [HA].[OH^-] /[A^-]

1.38 x 10^-4 = y²/(0.267-y)

y² = 1.38 x 10^-4 (0.267-y) M

y² + 1.38 x 10^-4 y - 3.6846 x 10^-5 = 0

On solving, we have

y = 0.006

Therefore, [OH^-] = y = 0.006 M

As,

pOH = - log [OH^-]

pOH = - log 0.006

pOH = 2.22

Also,

pH = 14 - pOH

So,

pH = 14 - 2.22

pH = 11.78

Hence, pH = 11.78

--------------------------------

2.

(a). Initial moles of HC₃H₅O₂ (acid) = Molarity x Volume in L

= 0.30 M x 0.200 L

= 0.06 mol

Similarly,

Initial moles of NaC₃H₅O₂ (salt) = Molarity x Volume in L

= 0.22 M x 0.300 L

= 0.066 mol

pH of buffer solution can be calculated by using Henderson-Hasselabalch equation :

pH = pK_a + log [Salt] /[Acid]

pH = - log K_a + log 0.066/0.06

pH = - log (1.3 x 10^-5) + log 1.1

pH = 4.89 + 0.0414

pH = 4.93

(b). After the addition of HI :

Initial number of moles of HI = 0.15 M x 0.010 L

= 0.0015 mol

Now, ICF table is :

-----------------------NaC₃H₅O₂..........+............HI -------------------> HC₃H₅O₂ .................+...................NaI

Initial .....................0.066 mol......................0.0015 mol................0.06 mol..............................................

Change.................-0.0015 mol...................-0.0015 mol...............+0.0015 mol........................................

Final ......................0.0645 mol........................0.0 mol...................0.0615 mol ....................................

Again,

pH = 4.89 + log 0.0645/0.0615

pH = 4.89 + 0.0207

pH = 4.91

Hence, pH becomes 4.91