What concentration of ammonia,[NH3],should be present in a solution with [NH4+]=

Question

What concentration of ammonia,[NH3],should be present in a solution with [NH4+]= 0.732 M to produce a buffer solution with pH=9.12? For NH3 Kb= 1.8Timesl0-5 A buffer solution of volume 500. mL contains 1.68gNH3 and 4.05g(NH4)2S04. Calculate the pH of the original buffer. If 0.88 g NaOH is added to the original buffer, what will be the pH? How many milliliters of 12 M HC1 must be added to the original buffer to change its pH to 9.00? If 100. mL of 1.50 M HC1 is added to the original buffer, what will be the pH? In the titration of 25.00 mL of 0.100 M HC2H302 with 0.200 M NaOH, determine the pH on the addition of 1.5 mL, 12.5 mL, 12.8 mL of NaOH. In the titration of 10.00 mL of 0.04050 MHCl with 0.01120 M Ba(OH)2 in the presence of the indicator 2,4-dinitrophenol, the solution changes from colorless to yellow when 17.90 mL of the base has been added. What is the approximate value of pKHin for 2,4-dinitrophenol? Is this a good indicator for the titration? Calculate the molar solubility of PbBr2(Ksp=4.0Times10-5 ) in pure water, 1.5 M KBr.

Explanation / Answer

(i) Ka =

becomes

Ka =

therefore: [H+(aq)] = Ka [acid(aq)] / [salt(aq)] mol dm-3, and taking -log10 of both sides gives

(iii) pHbuffer = -log10(Ka x [acid(aq)] / [salt(aq)])

or (iv) pHbuffer = pKa + -log10([acid(aq)] / [salt(aq)])

or (v) pHbuffer = pKa + log10([salt(aq)] / [acid(aq)])

hence we get,

for acid buffer

ph = pka + log[salt]/[acid]

(i) Kb =

becomes

Kb =

therefore: [OH-(aq)] = Kb [base(aq)] / [salt(aq)] mol dm-3, and taking -log10 of both sides gives

(iii) pOHbuffer = -log10(Kb [base(aq)] / [salt(aq))

or (iv) pOHbuffer = pKb + -log10([base(aq)] / [salt(aq)])

or (v) pOHbuffer = pKb+ log10([salt(aq)] / [base(aq)])

for basic buffer

poh = pkb + log[salt]/[base]

it is well known that pH + pOH = 14

a)

no of moles of nh4+ is = 4.05/132 as 132 is the molecular weight

concentration of nh4+ will be twice of that of (nh4)2so4

[salt] = [nh4+] = 2*(4.05/132)/volume of solution = 2*(4.05/132)/0.5 =0.122727(MULTIPLIED BY 2 AS EACH MOLECULE GIVE 2 NH4+)

[base] = [nh3] = 1.68 / 17 / 0.5 = 0.1976(as 17 is the molar mass of NH3 and volume is 0.5L

here base is NH3 and salt is NH4+ whose concentration are found out previosly,

poh = pkb of nh3 + log([salt]/[base])

= 5-log1.8 + log(0.122727/0.1976)

= 4.5378

pH = 14 - pOH

= 9.4621

b)0.88 g naoh = 0.022mol

NH4)2SO4 + 2NH4OH

this NaOH will react with (NH4)2SO4 AND FORM NH4OH which is base equivalent to NH3(actually NH3 + H2O) AND DESTROY NH4+ WHICH ACT AS SALT

[SALT] =   (2*(4.05/132) - 0.022)/0.5 = (0.06136 - 0.022)/0.5 = 0.03936/0.5 = 0.078728 (DIVIDING BY 0.5 AS IT IS 500ML WATER

[BASE] = (1.68 / 17 + 0.022)/0.5 = (0.0988+0.022)/0.5 = 0.1208/0.5 = 0.241

POH = 4.7447 -0.485887 = 4.2588

PH = 9.74

C)

as expected pH is 9, pOH = 4

SUBSTITUTING IN PREVIOUS EQUATION

4 = 4.7447 + LOG [SALT]/[ACID]

[SALT]/[ACID] = 1.8

HCl + NH4OH => NH4+ + Cl- (weak base)+ H2O

X -X 0.0988 - X     0.06136 + X X X

LET X MOLE BE ADDED

0.06136 + X = SALT

0.0988 - X = BASE PRESENT IN 0.5L OF SOLUTION

1.8 = 0.06136+X/0.0988-X

0.17785-0.06136 = 2.8X

X = 0.0416 MOLES

so 0.0416 moles og HCl is required

12 M HCL

Volume of Hcl required = No. of moles / molarity = 0.0416/12 = 0.003467 L = 3.4567 ml OF SOLUTION IS NEEDED

HENCE 3.467 ml OF SOLUTION IS NEEDED

D) 100ml OF 1.5M SOLUTION = 0.15 moles HCl is added to the solution

WHICH IS MORE THAN AMOUNT OF NH3 PRESENT . HENCE IT WILL CONSUME WHOLE NH3

HENCE NO MORE BUFFER

AMOUNT OF HCL PRESENT AFTER NH3 IS COONSUMED = 0.15 - 0.0988 = 0.0512 IN 500ml OF WATER

[HCL] = 0.0512/0.5 = 0.1024

HENCE [H+] = 0.1024

pH = -LOG[0.1024] = 1.07

11)

PKA = 4.7447 ACETIC ACID

25 ML 0.1 M CH3COOH = 2.5 mmol =>2.5mmol OF H+ IONS

i) 1.5ml NAOH SOL = 0.3mmol

AFTER REACTION NAOH WILL BE CONSUMED GIVING 0.3 mmol OF SALT AND 2.2mmol OF ACID

PUT IN THE FORMULA

ph = pka + log[salt]/[acid]

=4.7447 + LOG0.3/2.2 = 4.89

ii) 12.5 ML = 2.5m MOL

THEY WILL COMPLETELY REACT GIVING NO ACID AND COMPLETE 0.25 MOL OF CH3COONa

WHICH WILL UNDERGO HYDROLYSIS

FOR HYDROLYSIS

PH = 7+0.5*Pka= 9.37

iii)12.8 mL = 2.56mmol THIS WILL CONSUME WHOLE ACID AND 0.06 mmol OF BASE WILL BE LEFT IN 37.8 ML OF SOLUTON

[OH-] = 1.587 * 10-3

PH = 14-2.7993 = 11.2007

12)

12)

mmols = mili moles = 10-3 mols

NO OF MOLE HCL = molarity * volume = 0.405 mmol

THAT OF BA(OH)2 = 0.2 mmol=> 0.40096mmol OF OH-

HENCE 0.005 mmol OF HCl REMAINS

[H+] IN FINAL SOL = 0.00096/27.9 = 3.44 * 10-5

pH= 4.4633

pH of the solution was 4.4633 at the time of change in colour

now, all indicators are weak acid or base forming buffer while neutralizatio and the colour change is considerable when [salt]/[acid] = 3(salt is 0.75 and acid is 0.25)

(i) Ka =

becomes

(ii) Ka =

therefore: [H+(aq)] = Ka [acid(aq)] / [salt(aq)] mol dm-3, and taking -log10 of both sides gives

(iii) pHbuffer = -log10(Ka x [acid(aq)] / [salt(aq)])

or (iv) pHbuffer = pKa + -log10([acid(aq)] / [salt(aq)])

or (v) pHbuffer = pKa + log10([salt(aq)] / [acid(aq)])

pH = 4.4633 = pkHIN + log AS IT GIVES COLOR WHEN PH = pkHIN + log3

pkHIN = 4.4633-log3 = 3.9

(i) Ka =

[H+(aq)] [A-(aq)] ------------------------- mol dm-3       [HA(aq)]

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