Titration Calculation (2 Viewers)

[micuzzo]

Can someone pleas do this question...please...

Thank You

 

[Timothy.Siu]

28a) n=CV
C=n/V
=1.314/105.99/0.25=0.049589583moles per litre

b) umm
Na₂CO₃ + 2HCl --> 2NaCl+CO2+H20

M_aC_a=2M_bC_b

C_a=2M_bC_b/M_a
C_a=0.106...=0.106 (3dp)

 

[micuzzo]

28a) n=CV
C=n/V
=1.314/105.99/0.25=0.049589583moles per litre

b) umm
Na₂CO₃ + 2HCl --> 2NaCl+CO2+H20

M_aC_a=2M_bC_b

C_a=2M_bC_b/M_a
C_a=0.106...=0.106 (3dp)

i think the answer is 0.0938 mol/L ... i could b wrong... can someon re-try

 

[Timothy.Siu]

i think the answer is 0.0938 mol/L ... i could b wrong... can someon re-try

yeah thats right, i typed in my stuff wrong

C_a=(2x23.5x0.05)/25=0.0938

 

[micuzzo]

yeah thats right, i typed in my stuff wrong

C_a=(2x23.5x0.05)/25=0.0938

ahh k thank u dude

 

[chantszhin]

can i just ask...........
- do u have to divide the concentration / number of moles from 250mL to 23.45mL??? or does it stay the same unchanged throughout the whole thing.......

and i got 0.02325M...LOL quite off,,,,,failll

 

[brenton1987]

- do u have to divide the concentration / number of moles from 250mL to 23.45mL??? or does it stay the same unchanged throughout the whole thing

Concentration will never be divided by number of moles. Apart from that I dont understand what youre saying.