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Answer :
according to this formula:
PH= Pka + ㏒[A^-]/[HA]
when we have the value of PH=4.93 & Pka (missing in your question) = 4.76 &
no.of mole of acetate = 40 mmol & no.of mole of acetic acid= 60
so by substitution:
4.93 = 4.76 + ㏒((40+X)/(60-X))
0.17 = ㏒ ((40+X)/(60-X))
∴X= 19.66 mmol = 0.019 mol
finally we can get the volume from this formula of the molarity:
molarity = number of moles / volume
6 = 0.019 / V
∴V= 0.0032 mL = 3.2 L
PH= Pka + ㏒[A^-]/[HA]
when we have the value of PH=4.93 & Pka (missing in your question) = 4.76 &
no.of mole of acetate = 40 mmol & no.of mole of acetic acid= 60
so by substitution:
4.93 = 4.76 + ㏒((40+X)/(60-X))
0.17 = ㏒ ((40+X)/(60-X))
∴X= 19.66 mmol = 0.019 mol
finally we can get the volume from this formula of the molarity:
molarity = number of moles / volume
6 = 0.019 / V
∴V= 0.0032 mL = 3.2 L
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The Volume of NaOH is mathematically given as
V=3.2 L
Volume of NaOH
Generally the formula for PH is
Ph= Pka + log[A^-]/[HA]
4.93 = 4.76 + log((40+X)/(60-X))
0.17 = log ((40+X)/(60-X))
Therefore
X= 19.66 mmol = 0.019 mol
Generally the formula for molarity is
M= number of moles / volume
6 = 0.019 / V
V=3.2 L
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