please show step by step and circle each solution. It would really help me to un

Question

please show step by step and circle each solution. It would really help me to understand the concept when broken down. Thank you

Assurme you dissolve 0.105 g of the weak acid benzoic acid, C,Ht,CO,H, in enough water to make 1.00 x 10° ml of solution and then titrate the solution with 0.188 M NaOH K, for benzoic acid- 6.3 x 105) what was the pH of the original benzoic acid solution? pH- h. What are the concentratioes of all of the following ions at the equivalence point Na, HO, OH and C, H, Co,-7 What is the pH of the solation an the equivalence point? c. pH -

Explanation / Answer

(a)

Moles of solute = Mass/MW = 0.105/122 = 0.00086

Conc, C = moles/Volume = 0.00086/0.1 = 0.0086 M

Using Ostwald's dilution law we have:

Initial [H+] = (Ka*C)0.5 = (6.3*10-5*0.0086)0.5 = 7.36*10-4 M

So,

pH = 4-log(7.36) = 3.13

(b)

Using relation:

M1V1 = M2V2

Putting values:

0.0086*0.1 = 0.188*V2

V2 = 0.0046 L

So, conc of conjugate base at equivalence point is:

C' = moles/Volume = 0.00086/(0.0046+0.1) = 0.0082 M

At equivalence point, the conjugate base reacts with a molecule of water to produce the acid molecule back and a hydroxyl ion.

At equivalence point we have:

Using Ostwald's dilution law we have:

[OH-] = (Kb*C')0.5 = ((10-14/(6.3*10-5))*0.0082)0.5 = 1.14*10-6 M

[H+] = 10-14/[OH-] = 8.7*10-9 M

[Na+] = 0.0082 M

[CH3COO-] = C' - [OH-] = 0.0082 M approx.

(c)

Ae equivalence point we have:

pH = -log([H+]) = 9-log(8.7) = 8.06

Hope this helps !

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