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what is the pK b of a 0.450 M solution of hypoiodous acid, HOI? ( Ka1 = 2.3x10 -

ID: 894257 • Letter: W

Question

what is the pKb of a 0.450 M solution of hypoiodous acid, HOI? ( Ka1 = 2.3x10-11)

a. 10.29               b. 10.64               c. 5.15                d. 3.36                e. 8.85

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a solution is prepared by adding 0.10 mol of potasium acetate, KCH3COO, to 1.00 L of water. which statement about the solution is correct?

a. concentration of acetate ions > than the concentration of potassium ions

b. solution is neutral

c. concentration of potassiuom ions and acetate is identical

d. solution is basic

e. solution is acidic

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what is the pH of a 0.879 M solution of phosphoric acid, H3PO4? Ka1 = 7.5x10-3   Ka2 = 6.2x10-8   Ka3 = 4.8x10-13

a. 3.64               b. 7.28               c. 1.09               d. 2.18               e. 0.06

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what is the pH of a 0.900M solution of canic acid, HOCN? Ka1 = 3.5x10-4

a. 12.2               b.  1.8               c. 3.5               d. 10.5               e. 4.6

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what is the pOH of a 0.050 M HBr solution?

a.  1.30               b. 1.12               c.  12.70               d.  11.00               e. 0.89

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butyric acid is responsible for the odor in rancid butter. A solution of 0.25 M butyric acid has a pH of 2.71. what is the Ka for the acid?

given the equation Ag2CO3(s) <--> 2Ag+(aq)+CO32-(aq). what is the solubility product expression, Ksp, of silver carbonate.

Ksp silver carbonate = 8.4x10-12

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which of the following substances has the greatest solubility in water?

a. ZnCO3 (Ksp=1.2x10-10)         b. Ag2CO3 (Ksp=8.4x10-12)      c. Cr(OH)3 (Ksp=6.7x10-31)      d. AgCl (Ksp= 1.8x10-10)                   e. Mn(OH)2 (Ksp=2.1x10-13)

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what is the pH of a solution after titrating 50.0 mL of 0.650 pyridine, C5H5N (Kb = 1.8x10-9), with 50.0 mL of HCl?

a. 4.47               b. 2.71            c. 5.74                d. 2.87                  e. 5.43

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a 10.0 mL sample of 0.75 M CH3CH2CO2H (ka1=1.3x10-5) is titrated with 0.30 M NaOH. what is the pH of the solution after 22.0 mL of NaOH have been added to the acid?

a. 4.89      b. 5.00      c. 5.75     d. 4.12     e. 2.50

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what is the pH of a solution prepared with 0.50 M HC3H5O3 (lactic acid) (Ka1 = 1.4x10-4) and 0.75 M NaC3H5O3 (sodium lactate)?

a. 0.46     b. 2.00     c. 3.68     d. 4.03     e. 3.85

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calculate the solubility of sliver chromate , Ag2CrO4 (Ksp=1.1x10-12), in 0.005 M Na2CRO4.

a. 3.4x10-5     b. 1.0x10-6     c. 8.2x10-5     d. 5.2x10-7     e. 7.4x10-6

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if the solubility of barium fluoride is 7.2x10-3 M, what is the Ksp of BaF2?

a. 7.5x10-7     b. 3.7x10-7     c. 7.2x10-3     d. 1.5x10-6     e. 5.2x10-5

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which of the following aqueous misxtures would be a buffer system?

a. NH3(Kb=1.8x10-5), NaOH                   b. H2SO4(Ka1=very large)(Ka2=1.2x10-2), CH3COOH(Ka1=1.8x10-5)                           c. H3PO4(Ka1=7.5x10^-3)(Ka2=6.2s10^-8)(Ka3=4.8x10^-13), NaH2PO4         d. HCl, NaCl           e. HNO3,NaNO3

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How many grams of NaOBr should be added to 50.0 mL of 0.650 M of HOBr(Ka1=2.0x10-9) to make a buffer at pH 8.90

Explanation / Answer

Solutions :-

a solution is prepared by adding 0.10 mol of potasium acetate, KCH3COO, to 1.00 L of water. which statement about the solution is correct?

Solution :- potassium acetate is the conjugate base of the acetic acid therefore when it added to water then it forms the OH- ions

So the solution will be basic in nature

Therefore correct answer is option ‘d’

what is the pH of a 0.879 M solution of phosphoric acid, H3PO4? Ka1 = 7.5x10-3   Ka2 = 6.2x10-8   Ka3 = 4.8x10-13

Solution :- using the ka1 lets calculate the pH

Ka= [H2PO4^-] [H3O+]/[H3PO4]

7.5*10^-3 = [x][x]/[0.879-x]

7.5*10^-3 * 0.879-x = x^2

Solving for x we get

x= 0.0775

therefore pH = -log [H3O+]

                        = - log [0.0775]

                       = 1.1

So the correct answer is option ‘c’

what is the pH of a 0.900M solution of canic acid, HOCN? Ka1 = 3.5x10-4

Solution : -

Ka = [ H3O+] [ OCN^-] /[HOCN]

3.5*10^-4= [x][x]/[0.900 –x]

3.5*10^-4 * 0.900-x =x^2

X= 0.0176

pH= -log [H3O+]

pH= -log [0.0176]

pH= 1.8

So the correct answer is option ‘b’

what is the pOH of a 0.050 M HBr solution?

Solution :- HBr is the strong acid

Therefore

pH= -log [H+]

pH= -log [0.050]

pH= 1.30

pH+ pOH = 14

therefore pOH = 14- pH

                            = 14 -1.30

                            = 12.70

Therefore the correct answer is option ‘c’