paper 1 questions redo

H+ A-/ HA

Sulfurous acid = H2SO3

Dissassociates into H+ and HSO3- and then H+ and SO32-

Divide 0.0611/2 and plug into H+ = - log (10)(ph)

1. Calculate Ka 

2. Calculate pH before acid added suing Ka and initial concentration of A- and HA 

3. Calculate initial moles of HA, A- and Acid Added 

4. FInal moles of acid added = 0 

5. Calculate change in moles using change in moles of acid added (assume 1:1:1 ratio) 

6. Using final moles  calculate concentration of HA and A- 

7. Plug into H+ = K x HA/A - and then  work out final pH (pH = - Log (10) (h+) 

8. Final pH - Initial pH = Change in PH

9. answer = 0.08

1. Recognise that Burgundy mixture is made of ½ Na2CO2 and ½ Copper Sulfate 

2. Moles of Hcl = Moles of Burgundy Misture

3. ½ moles of HCL = ½ moles of Na2CO3 (base) 

4. Calculare moles of HCl 

5. Calculate moles of Na2CO3 (½ Hcl)

6. Scale up to 250 

7. Calculate mass of Na2Co3

8. Calculate mass of CuSO4 ( mass of burgundy - mass of Na2CO3)

9. Mass of Cu2+ ions by (mass of CUsO4 x Mr of Cu/ Mr od CuSO4)

   

if we use nomral method the mr of hydrated salt is less than moles of salt. We have tp divide scaled up moles by 2 because we found the moles of Al3+ but the formula unit is Al2 (SO4)3 so we need to divide by 2

1. moles of NaOH (A-)

2. Moles of HA (butanoic acid)

3. Moles of HA after A- added = Moles of NaOH - Moles of A-

4. Sub values into H+ = Ka x HA/A-

5. ph = -log(10)(h+)

mass of one mole of ions = mass x avogardo constant

balance atoms (S) first

add 2 lone pairs on oxygen

when doing gibbs calculaitons remember

gibbs energy eqaution = H - T (S/1000)

which values need to be X2

aomisaiton of chlorine (2 moleucles of chlrine be atomised)

Electron affinity of chlorine (2 moles of electron added to chlorine)

1. rank charges

2. rank size of cation

• same atomic number

• different mass number

ionisation energy = removing electrons (so more postive)

• remeber 2nd ionistion energy so write first ionisation energy and then 2nd

• NA— NA+ +e-

• Na—- Na2+ + e-

not the information in the quesiotn that we have been given

USe euqaiton C1V1=C2V2

Calculate H+ = Sqaure root KA x HA

Calcualte Ph

1. calculate Ph

2. Rearrnage Ka equation

   • remeber we are only squar rooting the numberator not the whole equation

   • TO reaarnge for H+ swap A- and H+

   • Calulate conc (plug values into equation

   • Calculate Moles (remember to half the volume)

   • Calculate mr

   • Calculate mass

• in excess replace all H2O ligands

• Change Charge to -3

• Remember to start from Al(H20)6 3+

• reacts as an acid in wate r

• Aq + H20 = aq (1 less h2o) (OH) i less charge + H3O+

1:2 initial mole ratio with H so H is not 0 but 2 x CO 

label acid and base in reaction

Produces HCl = misty fumes and NaHSO4

Remember Cl and any H most likely

H2SO4 is a stronger acid so NaCL is a base

A base (accoridng to bronsted lowry) accepts a proton

Br2 + SO2 + H2O

when Cl is added to NaBr yellow orange solution forms because Br2 is produced which is oraange bronw in water

Write reaction eqaution = displacement reaciton

Cl2 + NaBr ==== NaCl + Br2

Idenitfy OA and RA

Cl2 —— CL-

Br - ——— Br2

• balance atoms

• Balance H and O (not relevant)

• Balance charge

what are the standard conditons

Co (OH) and Co (OH)3

-0.76 = right/ reduced

use ph resulting solution tabkek

• calculate moles of HA

• HClO4 so 1 dissaosscairon of H+ so moles of HA = Moles of H+

• Calculate moles of A-

• Calculate moles of H+ in A- (hydroxixde)

• Calculate resulting moles of H+ = Moles of Acid - moles of H+ in Base

• Calculate Conc of H+

calculate moels of acid

Calculate moles of base

Calculate leftover moles of OH

H+ = kw/ OH (leftover moels)

higher ka = stornger acid

polarisis/ weaknens OH bond

Cl is highly electrongative so attracts electrons more strongly

Cis same side (next to or one aqay form eachother)

Trans - oppsoitve side

to draw trans make sure the cl is more than one away from eachthe r