When Justin added deionized water to the volumetric flask, he accidentally filled the flask so that the water level was slightly
1 answer:
You might be interested in
You missed a lot of details in your question, so when we have the complete question as the attached picture so, the answer would be:
when we have the value of Ka = 1.8 x 10^-5 so, we can use it to get the Pka value
by using this formula:
Pka = -㏒Ka
= -㏒(1.8 x 10^-5)
= 4.7
now, after we have got the Pka we need now to get moles of NaC2H3O2 and
moles of HC2H3O2:
when moles of NaOH = 0.015 moles
when moles NaC2H3O2 after adding NaOH
= initial mol NaC2H3O2 + mol NaOH
∴moles NaC2H3O2 = 0.1 + 0.015 = 0.115 moles
and moles HC2H3O2 after adding NaOH
= initial mol HC2H3O2 - mol NaOH
∴ moles HC2H3O2 = 0.1 - 0.015 = 0.085 moles
so, when we have moles [HC2H3O2] &[NaC2H3O2] so we can substitution its values in [A] &[HA] :
by using H-H equation we can get the PH:
when PH = Pka + ㏒[A]/[HA]
PH = 4.7 + ㏒0.115/0.085
= 4.8
when we have the value of Ka = 1.8 x 10^-5 so, we can use it to get the Pka value
by using this formula:
Pka = -㏒Ka
= -㏒(1.8 x 10^-5)
= 4.7
now, after we have got the Pka we need now to get moles of NaC2H3O2 and
moles of HC2H3O2:
when moles of NaOH = 0.015 moles
when moles NaC2H3O2 after adding NaOH
= initial mol NaC2H3O2 + mol NaOH
∴moles NaC2H3O2 = 0.1 + 0.015 = 0.115 moles
and moles HC2H3O2 after adding NaOH
= initial mol HC2H3O2 - mol NaOH
∴ moles HC2H3O2 = 0.1 - 0.015 = 0.085 moles
so, when we have moles [HC2H3O2] &[NaC2H3O2] so we can substitution its values in [A] &[HA] :
by using H-H equation we can get the PH:
when PH = Pka + ㏒[A]/[HA]
PH = 4.7 + ㏒0.115/0.085
= 4.8