Suppose that you wanted to prepare a acetate ion/acetic acid buffer solution wit

Question

Suppose that you wanted to prepare a acetate ion/acetic acid buffer solution with a pH of 4.35. What is the value 34 Marks: 2 off [A l/IHA) for the correct buffer solution for this event? The Ka for acetic acid is 1.8 x 10 5. a. 0.39 Choose one answer. b. 0.407 c. 0.372 d. 4.74 Suppose that you have a 43 mL solution of 0.23 M HCl that is being titrated with 0.17 M NaOH. You stop the 35 titration after adding 27 mL of NaOH. What is the change in pH atthis point? Marks: 2 a. 0.27 Choose one answer. b. 0.48 c. 1.12 Suppose that we are looking at the titration of acetic acid with NaOH. At the equivalence point, we have 0.09 36 Marks: 2 moles of acetate (CH3Coo At this point, the following reaction occurs and reaches equilibrium: CH3COO (aa) H2O(e) CH3COOHaa) OH (aa), where Kb is 5.6 x 10 What is the pH of the solution? a. 7.1 Choose one answer. b. 3.7 c. 7.0 d. 8.85

Explanation / Answer

(34) Apply Handersson's equation to get the desired answer-

pH = PKa + log [salt] / [ Acid ]

= -log Ka + log [ Salt ] / [ Acid ]

   4.35 = - log ( 1.8 x !0-5  ) + log { [Salt ] / [ Acid ] }

= 4.7447 + log{ [Salt ] / [ Acid ] }

So, log { [Salt ] / [ Acid ] } = - 4.7447 + 4.35

= - 0.3947

& [Salt ] / [ Acid ] = antilog ( -0.3947 )

= 0.4069

hence, the correct answer is -----------------> " b''

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(35 ) Find out the volume of HCl solution used by appplying relation-

0.23 x V = 0.17 x 27

V = ( 0.17 x 27) / 0.23

= 19.95 ml

So volume of 0.23 M NaOH used for titration = 19.95 ml which is also equivalent to 19.95 ml. of 0.23M HCl soln.

Volume of HCl left = 43- 19.95 = 23.05 ml of 0.23M HCl soln

but now this volume is present in (43 + 27 ) ie. 70 ml of soln. , hence calculate the molarity of this soln. & find out pH

70 x M = 23.05 x 0.23

= 0.07573M

& pH = - log ( 0.07573 )

= 1.12

hence, the correct answeer is--------------- ( c )

------------------------------------------------------------------------------------------------------------------------------- Pl. post the other question separately, for answering, Glad to help.

However , here is the answer-

(36) Study the equilibria & find out [OH- ] as below-

CH3 COO^- (aq) + H2 O (l) <---------------> CH3COOH (aq) + OH^-(aq)

initial conc. (moles) -------------- 0.09-----------------------------------------------o ----------------0

change ------------------------------ -x ------- -x ----------------------------------- +x ---------------+x

at equilibria ---------------------------(0.09- x ) H2O --------------------------------x ----------------- x

K = { [CH3 COOH] [ OH^- ] } / [ CH3COO^- ]

= (x) (x) / (0.09 - x )

= x^2  / 0.09 (x is usually so small & negligible, hence is ignored frrom the denominator)

Again , K = x^2 / 0.09   = Kw / Ka = 10-14 / 1.8 x 10-5  *

So, x^2 = 49.95 x 10^-12  

x = 7.07 x 10^-6  

This represents the [ OH- ] also ie. [OH- ] =   7.07 x 10-6

Again [ H+] [ OH- ] = 10 -14   

therefore, [H+ ] = 10-14 / 7.706 x 10-6  = 1,4 x 10-9

& pH = - log[H+] = - log(1.4 x 10-9 )

= 8.85

Thus the correct answer is ------------------------ (d)

Note-

* The value of Ka  is  calculated as 1.8 x 10-5   using the relation-

pKa    + pKb   = 14

pKb   is given , we can calculate pKa

Known to us pKa value = - log Ka

  

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