7) Suppose you have 0.10 M aqueous solutions of each of the following substances

Question

7) Suppose you have 0.10 M aqueous solutions of each of the following substances: HINO, KCI, NH4CI, KOH, and NaC2H3?2. Place these solutions in order of increasing pH, that is, from the lowest pH to the highest pH a) HNO3, KCI, NH4CI, KOH, NaC2H:02 b) KCl, NH4Cl, HNO3, KOH, NaC2H3O2 c) HNO3, NH4CI, KCI, NaC2H302, KOH d) NH4CI, HNO3, KCI, KOH, NaC2Hs02 e) HNO3, NaC2H3O2, KCI, NH&CI;, KOH 8) Find the percent ionization of a 0.25 M HCN solution. The Ka for HCN is 6.4 × 10-5 A) 1.6% B) 1.4 x 10-2 % C) 3.6 % D) 2.5 x 10-2 % E)47% 9) Determine the Ka of an acid whose 0.294 M solution has a pH of 2.80. A) 1.2 x 10-5 B) 8.5 x 10-6 C) 2.7 D) 4.9 x 10-7 E) 5.4 x 10-3

Explanation / Answer

HNO3 is strong acid . For strong acids PH value 0 to 3

NH4Cl is acidic nature due to cationic hydrolysis. NH4^+(aq) + H2O --------------> NH3(aq) + H3O^+ (aq).

The NH4Cl solution Ph value less than 7

KCl is salt of strong acid and strong base . So PH = 7

NaC2H3O2 is basic nature due to anionic hydrolysis . C2H3O2^- (aq) + H2O -----------> HC2H3O2(aq) + OH^- (aq)

The NaC2H3O2 solution PH value greater than 7

KOH is strong base . For strong bases PH value 10 to 14

HNO3, NH4Cl, KCl, NaC2H3O2 , KOH

c. HNO3, NH4Cl, KCl, NaC2H3O2 , KOH >>>>>>answer

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