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A solution of NaOH(aq) contains 5.4 g of NaOH(s) per 100.0 mL of solution. Calculate the pH and the pOH of the solution at 25 degreeC. Assuming complete dissociation, what is the pH of a 4.40 mg/L Ba(OH)2 solution? If the Ka of a monoprotic weak acid is 6.8 Times 10-6, what is the pH of a 0.42 M solution of this add? The Ka value for acetic acid, CH3COOH(ag), is 1.8 Times 10-5. Calculate the pH of a 1.80 M acetic acid solution. Calculate the pH of the resulting solution when 3.00 mL of the 1.80 M acetic acid is diluted to make a 250.0 mL solution. Enough of a monoprotic acid is dissolved in water to produce a 0.0129 M solution. The pH of the resulting solution is 2.56. Calculate the Ka for the acid. The Ka of a monoprotic weak acid is 6.74 Times 10-3. What is the percent ionization of a 0.173 M solution of this add? If the Kb of a weak base is 7.2 Times 10-6, what is the pH of a 0.19 M solution of this base? A certain weak base has a Kb of 9.00 Times 10-7. What concentration of this base will produce a pH of 10.35? For the diprotic weak add H2A, Ka1 = 3.1 Times 10-5 and Ka2 = 5.1 Times 10-7. What is the pH of a 0.0500 M solution of H2A? What are the equilibrium concentrations of H2A and A2- in this solution? NH3 is a weak base (Kb = 1.8 Times 10-5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.078 M in NH4Cl?

Explanation / Answer

1) NaOH moles = 5.4/40 = 0.135

NaOH is strong base and dissociates completly , hence OH- moles = 0.135

vol =100 ml = 0.1L , hence [OH-] = 0.135/0.1 = 1.35 , pOH = -log ( 1.35) = -0.13

pH = 14-(-0.13) = 14.13

2) 4.4 mg/L = 0.0044 gm/L , Ba(OH)2 Molarity = 0.0044/171.34 = 2.568 x 10^-5 M

( where 171.34 gm/mol is mol wt of Ba(OH)2 )

[OH-] = 2 x 2.568 x 10^-5 = 5.136 x 10^-5 , pOH = -log [OH-] = -log ( 5.136x10^-5) = 4.29

pH = 14-4.29 = 9.71

3) HA <--------> H+ + A-

at equi [HA] = 0.42-x , [H+]=[A-] =x ,

Ka = 6.8 x 10^-6 =[H+][A-]/[HA]

6.8 x 10^-6 = x^2/( 0.42-x)

x = [H+] = 0.0017 , pH = -log ( 0.0017) = 2.77

4) CH3COOH <-----> CH3COO- + H+

at equi [CH3COOH] = ( 1.8-x) , [CH3COO-]=[H+] =x,

K = 1.8 x 10^-5 = ( x^2/1.8-x)

x = [H+] = 0.0057, pH = -log ( 0.0057) = 2.24

(ii) moles of CH3COOH = 1.8 x 3/1000 = 0.0054 , vol = 0.25 L ,

1.8 x 10^-5 = ( x/0.25)^2 / ( 0.0054-x/0.25)

x = 0.000154=H+ moles , [H+] = ( 0.000154/0.25) = 0.000616 , pH = -log ( 0.000616) = 3.2

5) [HA]= 0.0129, [H+ ] = 10^-2.56 = 0.002754

hence Ka = [H+][A-]/[HA] = ( 0.002754^2) /( 0.0129-0.002754) = 0.000748 = 7.48 x 10^-4

6) Ka = 6.74 x 10^-3 , [HA] =0.173,

Ka = 6.74 x10^-3 = ( x^2/0.173-x) , x=[H+]=[A-] = 0.03094

% ionization = 100 x [A-]/[HA] = 100 x 0.03094/0.173 = 17.88 %

7) BOH <----> B+ + OH-

Kb = 7.2 x 10^-6 = x^2/( 0.19-x)

x =[OH-] = 0.001166 , pOH = 2.93 , pH = 14-2.93 = 11.07

8) pH = 10.35, pOh = 3.65, [OH-] = 10^-3.65 = 2.24 x 10^-4 , Kb = 9x10^-7

Kb = 9x10^-7 = ( 2.24x10^-4)^2/( M-2.24 x10^-4)

M = 0.0554 is conc of base

9) Ka1 = 3.1x10^-5 , Ka2 = 5.1x10^-7 , [H2A]=0.05

at equ Ka1 = [H+][HA-]/[H2A]

3.1x10^-5 = x^2/( 0.05-x)

x = [HA-] = 0.00123

ka2 = [H+}[A2-]/[H2A]

5.1x10^-7 = x^2/( 0.00123-x)

x =[A2-] = 2.48 x10^-5 M

total [H+] = (0.00123+0.0000248) = 0.00126 , pH = 2.9

[H2A] = ( 0.05-0.000123) = 0.0488 M

10) N44Cl = 0.0178 M

at NH4+ <-> NH3 + H+

Ka = Kw/Kb = 10^-14/1.8x10^-5 =5.556 x 10^-10 = x^2/( 0.0178-x)

x=[H+] = 3.146 x10^-6 , pH = 5.5

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