Question

Hey! Can you please help me with the following two questions I am highly confused and I don't really understand them at all. Thank you!! 16, Please help me with them, we have a test this thursday and problems that are similar are supposed to be on there. Thank you so much!!!

1.The pH of a 0.14-M solution of phthalic acid (H₂C₈H₄O₄) is measured to be 1.91. Use this information to determine a value of K_a for phthalic acid.

• H₂C₈H₄O₄(aq) + H₂O(l) HC₈H₄O₄^-(aq) + H₃O⁺(aq) K_a =______

2.What is the pH of a 0.142 M aqueous solution of sodium acetate, NaCH₃COO? (K_a for CH₃COOH = 1.8×10^-5)

pH =______

Answer 1

ANSWER is Given: - Comentration of plethale and = 0.14 pH = 1.91 To find : ka=9 solution : H₂ CG H₂O4 (aq) + H₂O(e) H Ca Hay au rag) + H 30 tags Initial comen (M) 0.14. change in cone (M) Equilibrium cone (M) O O X 0.14-1 n Now we know, pH = -log [H3077 tog Enzot] 10-1.91 1.91 Cu304] 0.01230 =k Now applying (4307 [4 Cg My On ka to the above equation ka [H2 C&H 404 n² 14 0.14-1 (0.01230) 2 2 0.0001512 0.14 0.01230 014-X 0.0001512 0.001184 0.1277 ka 0.001184

To find 0.142M aqueous solution of soduin aulat. Ka for Chaloon 1.88105 - pred solution: sodun acetate is the salt of weak and and strong base so Cfg coola Also, CH₂600 +4₂0 CH₂ 2004 +0H a + NaT Also know Kakb Kako kb so kw 1x1024 1810-14 1.8810-5 5.555 x1010 tou 0 tu th 0.142-1 и n 1 kb = Now CH₂ LOO THEO CH₂COOH Initial concentration 0.142 change in com() equilibrum so kb = [CM3won] Cor [CM360o 5.555 X160 (As concentration of 0.142-1 [3 coo] = [or] =n x² (5-555 X1070) 0.142-4 But, sime the value of 이 kb & nuky small asume the nealue of of n is small neglecting so will take 0.142 x2 n so

so x = N 0.142(5.55X10-10) N 7.881810-11 8.8819 x 10-6 x = 8.8819 X10-6 x = Гон1 Also, we know pH + poH = 14 pH = 14-pon = P4-[-log(om)] = 14-[-log (8.8819 810) = 14- 5.051494 PH z 8.94 0.142 M aqueous solution of sodium autate 8.94 pH of