A CERTAIN INDICATOR, Hin, HAS Ka=1.0 times 10^5. Hin HAS COLOR 1 AND In\' HAS CO

Question

A CERTAIN INDICATOR, Hin, HAS Ka=1.0 times 10^5. Hin HAS COLOR 1 AND In' HAS COLOR 2. WHICH IS TRUE IN A SOLUTION OF pH = 7.00 IT IS COLORLESS IT HAS COLOR 2 IT HAS COLOR 1 IT HAS COLOR1+COLOR2 FOR THE NEXT 4 PROBLEMS CONSIDERTHE TITRATION OF 20.00mL OF HCl(aq) WITH M=.400M BY .800M KOH(aq) DETERMINE THE pH Of THE MIXTURE BEFORE THE ADDITION OF ANY OH' 1.000. 0.398 0.300 NEED THE K_a OF THE HCl DETERMINE THE pH AFTER THE ADDITION OF 5.00 mLOH' 1.301 0.796 0.495 4.301 DETERMINE THE mL KOH NEEDED AT THE EQUIVALENCE PT. AND THE pH UPON THE ADDITION OF THIS VOLUME OF KOH. 10.0,7.00 20.0, 7.00 10.0,14.00 20.0, 3.50 DETERMINE THE pH AFTER THE ADDITION OF 20.00mL OH'. 0.70 10.699 11.85 13.30

Explanation / Answer

7) B) color 2

Exp : Ka = [HIn] / [H+][In-], Then If pH > pKa it will be ionized form [In-], and if pH < pKa it will be neutral form [HIn]. Here pKa = 5, pH = 7 so it will exist as anion [In-]

8) B) 0.398

Exp : pH = -log[H+] => -log[0.4] = 0.398

9) B) 0.796

Exp : 20 mL of 0.4 HCl (= 8 mmol) reacts with 5 mL of 0.8M of KOH (= 4 mmol). Hence 4 mmol of HCl remains in 25 mL solution. Therefor [H+] = 4/25 = 0.16 M, pH = -log[0.16] = 0.796

10) A) 10 mL, pH=7

for complete neutralization(pH = 7) 8 mmol of KOH needed, hence 10 mL of 0.8 M = 8 mmol

11) D) 13.30

Exp : 20 mL of 0.4 HCl (= 8 mmol) reacts with 20 mL of 0.8M of KOH (= 16 mmol). Hence 8 mmol of KOH remains in 40 mL solution. Therefor [OH-] = 8/40 = 0.2 M, pOH = -log[0.2] = 0.699, pH + pOH = 14, pH = 14 - 0.699 = 13.30

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