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KISI-KISI SOAL Subject : Chemistry Grade/semester : XI/2Time allocation : 4 x 45 minutes

Standard competence : Understanding the acid-base properties, measuring method and the application Basic Competence : Calculating amount of reactant and product in electrolyte solution from the result of acid-base titration.

No Indicator Question Answer Domain1 Calculating mole of a

substances, volume, and also concentration of solution.

In order to determine 0,1 M solution of Ca(OH)2, how much Ca(OH)2 required to be dissolved in a 500 ml volumetric flask?

Volume = 500 mlMolarity = 0,1 MMw = 74 g/molMass= ….?

M=mass x1000Mw x v

0,1 M= mass x100074 g/mol x 500 ml

0,1M x 74 g/mol x 500 ml = mass x 1000 3700 gram = mass x 1000 3,7 gram = massSo, we need 3,7 grams of Ca(OH)2

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2 Explaining the way to make certain molarity solution.

For an experiment, a student needs 300 ml of HNO3 solution 0,25 M. In the laboratory available 2.0 M solution of HNO3. Explain how does he get the solution he wants, if in laboratory available volumetric flask in volume 250 ml, 500 ml, and 1000 ml?.

Initial molarity = 2,0 MFinal molarity = 0,25 M in 300 ml Used the volumetric flask in volume 500 ml to get 300 ml of 0, 25 M HNO3 with once diluting process.

M1. V1 = M2 . V22 M . V1 = 0,25 M . 500 ml

2 M. V1 = 125 M. ml

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No Indicator Question Answer DomainV1 = 125 M. ml/ 2 M

V1 = 62,5 ml. So, we can put 62,5ml of 2,0 M solution to make 500 ml of 0,25 M HNO3. And we can put 300 ml of this solution by using volumetric pipette and beaker glass or graduated cylinder.

3 Calculating mole of a substance, volume and also concentration of solution

Determine the [H+] of 1,8 ml of concentrated H2SO4 98% with density 1,8 Kg/L dissolved in 200 ml of water!

Initial Volume of H2SO4 = 1,8 mlFinal Volume of H2SO4 = 200 ml% mass = 98%ρ = 1,8 kg/L Mr = 98

[H2SO4initial] = 1,8 x 10 x98

98[H2SO4initial] = 18 M

M1. V1 = M2 . V2 18 M. 1,8 ml = M2. 200 ml M2 = 0,162 M [H+] = 2x [H2SO] = 2 x 0,162M = 0,324 M

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A student has 10 ml of vinegar solution 0,1 M. It will be dissolved to be 1 liter solution. Determine the pH of this vinegar solution before and after diluting process! (Ka= 1x10-5)

Initial Volume of CH3COOH = 10 ml[CH3COOH] initial = 0,1 MFinal Volume of CH3COOH = 1 liter = 1000 ml.pH of CH3COOH before diluted:[H+] = √ka . Ma = √10−5 x0,1 = √10−6

= 10-3

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No Indicator Question Answer DomainpH = - log [H+] = - log [10-3] = 3.pH of CH3COOH after diluted:

M1. V1 = M2 . V2 0,1M. 10 ml = M2. 1000 ml M2 = 0,001 M[H+] = √ka . Ma = √10−5 x0,001 = √10−8

= 10-4

pH = - log [H+] = - log [10-4] = 4


There is 100 ml of NH4OH solution with pH 10. How many mole of that solution?(Kb = 10-5)

Volume of NH4OH = 100 ml = 0,1 LpH = 10pOH = 14-10 = 4[OH-] = 10-4

[OH-] = √kb . Mb10-4 = √10−5 x Mb Mb = 10-8/ 10-5

= 10-3MMb = mole / volume10-3M = mole / 0,1 LMole = 10-2 Mole

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6 Calculating the number of reactant required or product obtained from

Determine the amount of Calcium metal required to displace sodium in its 20 ml of 0,15 M sodium sulphate

It is metal displacement reactionMmol Na2SO4 = 20 ml.0,15 M = 3 mmol Ca + Na2SO4 → CaSO4 + 2Na

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No Indicator Question Answer Domainelectrolyte solution reaction.

solution! mol Ca = mmol Na2SO4

= 3 mmol = 0,003 mol Mass of Ca = mole Ca x Mr.Ca = 0,003 mol x 40 gr/mol = 0,12 grams

7 Identifying the kinds of chemical reaction in electrolyte solution.

Look at this picture:

a. What kinds of this reaction?b. Why the reaction can be

occurred?c. Calculate the concentration of the

product!

a. It is a is metal displacement reaction b. The reaction can be occurred because the

displacing metal or Mg is in the left side of Hydrogen in HCl in volta series.

c. Mol Mg = mass/ Mw Mg = 4 gr/24 gr mol-1

= 0,16 molMol HCl = 10 ml x 0,4 M = 4 mmol = 0,004 mol Mg + 2 HCl → MgCl2 + H2

0,16mol 0,004 mol0,002mol 0,004 mol 0,002 mol0,158mol - 0,002 mol

[MgCl2] = 0,002 mol/0,01L = 0,2 M

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8 Calculating the number of reactant required or product obtained from electrolyte solution reaction.

There is a reaction of acid oxide and baseSO3(g) + Ba(OH)2 → BaSO4(aq) + H2O(aq)Determine the mass of salt are formed from 1L Ba(OH)2 pH 13 and 4,48 liter SO3 gas (STP)!

pH = 13pOH = 14-13 = 1[OH-] = 10-1M[Ba(OH)2] = ½ x 0,1 M = 0,05 MMol Ba(OH)2 = 1 L x 0,05 M = 0,05 molMol SO3 = 4,48/22,4 = 0,2 mol SO3(g) + Ba(OH)2 → BaSO4(aq) + H2O(aq)

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4,0 gr of Mg metal

m 10 ml of 0,4M HCl

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No Indicator Question Answer Domain 0,2 mol 0,05 mol 0,05 mol 0,05 mol 0,05 mol 0,15 mol - 0,05 molMass of BaSO4 = mole x Mr BaSO4

= 0,05 mole x 223 = 11,65 grams9 Identifying the kinds of

chemical reaction in electrolyte solution

Preparation of Silver bromide for the manufacture of black and white film in industry is a metathesis reaction between silver nitrate and calcium bromide.a. Write the reaction equation!b. Determine the volume of volume

of silver nitrate 0,2M that will completely react with 70 ml of 0,2M calcium bromide!

c. How many particles of silver bromide are formed?

a. The reaction equation :2AgNO3(aq) + CaBr2(aq) → 2AgBr(s) + Ca(NO3)2(aq)

b.[CaBr2] = 0,2 MVolume CaBr2 = 70 mlMmol CaBr2 = M. V = 0,2 M. 70 ml = 14 mmolMmol AgNO3= 2 x mmol CaBr2

= 2 x 14 mmol = 28 mmolVolume AgNO3 = mmol/ M = 28 mmol/ 0,2M = 140 ml

c. Mmol AgBr = mmol AgNO3

= 28 mmol = 0,028 molNumber of particles = mol x 6,02. 1023

= 0,028 x 6,02. 1023

= 16,9 x 1021 molecules

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10 Calculating the number of reactant required or product obtained from

50 ml of 0,1 NaOH mixed with 60 ml of 0,1M of HCl. Determine the pH of this solution. If it isn’t neutral

Mmol NaOH = 50 ml . 0,1M = 5 mmolMmol HCl = 60 ml . 0,1M

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No Indicator Question Answer Domainelectrolyte solution reaction.

solution, how do you neutralize this solution if in the laboratory available H2SO4 0,05M and Ca(OH)2 0,1 M?

= 6 mmol The reaction: NaOH + HCl → NaCl + H2Oi: 5 mmol 6 mmol r: 5 mmol 5 mmol 5 mmol r : - 1 mmol 5 mmol[H+] = mmol/v total = 1mmol/ 110 ml = 0,009MpH = - log 9 x 10-3

= 3 – log 9 (acid properties)It can be neutralized with Ca(OH)2

2HCl + Ca(OH)2 → CaCl2 + 2H2OMmol HCl must equal with mmol Ca(OH)2

Mmol Ca(OH)2 = ½ mmol HCl = ½ . 1 mmol = 0,5 mmolV Ca(OH)2 = mmol / M = 0,5 mmol/ 0,1M = 5 ml.


8,5 grams of mixture magnesium metal and iron metal react with copper(III) sulphate. Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)2Fe(s) + 3CuSO4(aq) →Fe2(SO4)3(aq) + 3Cu(s)

Mass Mg + Fe = 8,5 gramsMass of Mg = xMole Mg = x/24Mass of Fe = 8,5 – xMole Fe = (8,5 -x)/56 Mg + CuSO4 → MgSO4 + CuMole Cu = mole Mg

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No Indicator Question Answer DomainFrom the reaction we get 15, 875 grams of copper precipitate.Determine the mass of each metal as reactant!

= x/242Fe + 3CuSO4 → Fe2(SO4)3 + 3CuMole Cu = 3/2 mole Fe = 3/2 ((8,5 -x)/56) = (25,5 – 3x)/ 112Mole Cu = x/24 + (25,5 – 3x)/ 11215,875/63,5 = x/24 + (25,5 – 3x)/ 1120,25 x 24x112= 112x + 612 – 72x672 = 612 + 40x60 = 40 xx = 1,5 gr. (mass of Mg)mass of Fe = 8,5 – 1,5 = 7 grams


0,56 grams of M metal react with 50 ml HCl produce 0,05 mole of H2

gas and MCl2 solution. a. Determine the atomic mass of M

metal and the concentration of HCl solution!

b. If in certain Pressure and temperature 2 liter of CO has mass 0,28 grams. Determine the volume of H2!

Mass of M = 0,56 gramsVolume HCl = 50 ml M + 2HCl → MCl2 + H2

a. Mole of H2 = 0,05 mole Mole of H2 = mole M = ½ mole of HClMole M = 0,005 moleAr M = mass/ mole = 56/0,05 = 112Mole HCl = 2 x mole H2

= 2 x 0,05 = 0,1 mole [HCl] = mole/v = 0,1 mole/0,05L = 2 M

b.V CO = 2 litersn CO = 0,28 gram/28 = 0,01 molen H2 = 0,05 molev Co/ v H2 = nCo/n H2

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No Indicator Question Answer Domain2/ v H2 = 0,01/0,052 x 0,05 = 0,01 x v H2

0,1 / 0,01 = v H2

10 Liters = v H2 13 Calculating the number of

reactant required or product obtained from electrolyte solution reaction.

mixture contains of 40% calcium carbonate and 60% calcium hydroxide. Determine the volume of 2 M hydrogen chloride to dissolve 25 grams of mixture

CaCO3 + CaOH2 = 25 gramMass CaCO3 = 40% x 25 = 10 gramsMole CaCO3 = 10/100 = 0,1 moleMass Ca(OH)2 = 60% x 25 = 15 grams Mole Ca(OH)2 = 15/58 = 0,26 moleCaCO3 + 2HCl → CaCl2 + CO2 + H2OMole HCl = ½ mole CaCO3

= ½ 0,1mole = 0,05 moleCa(OH)2 + 2HCl → CaCl2 + 2H2OMole HCl = ½ mole Ca(OH)2

= ½ 0,26 = 0,13 moleVolume HCl = 0,05/2M + 0,13/2 = 0,025 + 0,065 = 0,09 L = 90 ml.

14 Doing the acid-base titration.

Explain why volumetric pipette and burette are rinsed with a little of the solution before being used to measure out volume the solution?

It in order to the volume of the solution measured exactly.

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15 Calculating the concentration acid-base solution from the

Consider titration of 100 ml of 0,1 M HNO3 with 0,1 M of KOHa. How many milliliters of KOH are

V HNO3 = 100 ml[HNO3] = 0,1M[KOH] = 0,1 M

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No Indicator Question Answer Domainresult of acid-base titration.

required to reach the equivalence point?

b. What is the pH at reach the equivalence point?

c. Sketch the general shape of titration curve!

a. To reach the equivalence point the acid and base must be reacted completely. HNO3 + KOH → KNO3 + H2OBecause the reaction coefficient of HNO3 and KOH are same. So the mole of KOH = mole HNO3

[KOH]. Vol KOH = [HNO3]. Vol HNO3

0,1 M. vol KOH = 0,1 M. 100 ml Vol KOH = 10mmol/0,1M = 100ml

b. It is a titration between strong acid and strong base. [H+] = [OH-] = √ Kw[H+] = √10−14

= 10-7

pH = - log [H+] = 7

c. The curve:

16 Calculating the concentration acid-base solution from the result of acid-base titration.

Consider 80 ml of sulphuric acid can be neutralized by 50 ml of potassium hydroxide 0,2 M. Determine the mass of sulphuric acid that occurred in every liter of solution! (Mr H2SO4 = 98)

Volume H2SO4 = 80 mlVolume KOH = 50 ml [KOH] = 0,2 MMass of H2SO4…?The pH = neutral, so the reactant react completely H2SO4 + 2KOH → K2SO4 + 2H2O

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No Indicator Question Answer DomainMmol H2SO4 = ½ mmol KOH80 ml. [H2SO4] = ½ 50 ml. 0,2M[H2SO4] = 5 mmol/ 80 ml = 0,0625MMol H2SO4 = M. V = 0,00625M. 80 ml = 5 mmol = 0,005 molMass = mol x Mr = 0,005 x 98 = 0,49 grams


There are titration results of 20 ml of H2SO4 solution with 0,1 M of NaOH bellow:

TitrationVolume of

NaOH added1 10,1 ml2 10,0 ml3 9,9 ml

a. Write the reaction equation!b. Calculate the molarity of acid

that reacted!c. What is indicator that used in

titration!

V H2SO4 = 20 ml[NaOH] = 0,1 M

a. H2SO4 + 2NaOH → Na2SO4 + 2H2Ob. The molarity of H2SO4 in first titration

Ma.VaMb. Vb

=ab

Ma .20 ml0,1 M .10,1 ml

=12

40 ml x Ma = 1,01 mmolMa = 0,0252The molarity of H2SO4 in second titration

Ma.VaMb. Vb

=ab

Ma .20 ml0,1 M .10 ml

=12

40 ml x Ma = 1,0 mmolMa = 0,0250The molarity of H2SO4 in third titration

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No Indicator Question Answer DomainMa.VaMb. Vb

=ab

Ma .20 ml0,1 M .9,9 ml

=12

40 ml x Ma = 0,99 mmolMa = 0,0248M H2SO4 = (M1 + M2 + M3)/3 = (0,0252+ 0,0250 + 0,0248)/3 = 0,025 M

c. Indicator that used is phenolphthalein because it is a titration between strong acid and strong base with pH range between 8,2 – 10,0


100 ml of acetic acid (Ka= 10-5) 0,1 M is titrated by 100 ml NaOH . if in the reaction the base is remained and the pH of mixture is 12. Determine the mass of NaOH that added!

V CH3COOH = 100 ml[CH3COOH] = 0,1 MV NaOH = 100 ml pH mixture = 12POH = 2[OH-] = 10-2 M = M NaOH CH3COOH + NaOH → CH3COONa + H2ONaOH is remainedMmol NaOH remain = V total x M NaOH = 200 x 10-2

= 2 mmolMass NaOH = 2 mmol x Mr = 2 mmol x 40 = 80 g

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