I need help with these three problems.. show as much work as you can please. tha

Question

I need help with these three problems.. show as much work as you can please. thank you!

TLIE y THE HENDERSON-HASSELBALCH EQN CONSIDER THE BUFFER , 100M C2H30i/,200M HC2HsO2, Ka-1.8X10-5 WHICH IS THE CORRECT REACTION UPON THE ADDITION OF OH TO THE BUFFER? A) OH-(aq) + HC3H3O2(aq) > C2H3O1(aq) + H2O(I) C) OH (aq)H(aq) H20 (1) A) 4.74 ) 4,44 C) 5.04 D)NEED ITS VOLUME A) 4.37 B) 4.07 C) 5.22 D) 2.00 CONSIDER THE BUFFER IN #4. DETERMINE ITS pH CONSIDER THE SAME BUFFER ABOVE IF .0500 MOLH IS ADDED TO 5.00L OF THE BUFFER CALCULATE THE NEW pH? A CERTAIN INDICATOR HTn

Explanation / Answer

Q1: Answer: A

OH- ions are neutralized by the acetic acid present in the buffer solution.

Q2: Answer: B) 4.44

pH = pKa + log ([Salt] / [Acid])

Here, pKa = - log Ka = - log (1.8 x 10-5) = 4.745

pH = 4.745 + log (0.1/0.2) = 4.745 - 0.301 = 4.444

Q3: Answer: A) 4.37

In 5.00L buffer 0.0500 MOL H+ is added, which means for each litre of buffer 0.01 mol H+ reacted.

H+ will convert C2H3O2- into HC2H3O2,

H+ + C2H3O2- ----> HC2H3O2

0.01 mol H+ is added per litre, so 0.01 mol acetic acid concentration will increase and the corresponding salt will decrease by 0.01 mol.

pH = pKa + log ([Salt] / [Acid])

pH = 4.745 + log (0.09/0.21) = 4.745 - 0.368 = 4.377

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