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0.20 mol of benzoic acid and 0.10 mol of sodium benzoate are dissolved in 1L pure water. What is the pH of this solution? Given Ka of benzoic acid = 6.3×10^-5. Solve this using the Henderson-Hasselbalch Equation pls?
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Nadeeja Rathnayaka
0.20 mol of benzoic acid and 0.10 mol of sodium benzoate are dissolved in 1L pure water. What is the pH of this solution? Given Ka of benzoic acid = 6.3×10^-5. Solve this using the Henderson-Hasselbalch Equation pls?
Henderson-Hasslebalch Equation: pH = pKa + log([salt]/[acid])
pKa = -log(6.3*10^(-5)) = 4.20
pH = 4.20 + log(0.10/0.20) = 4.20 - 0.30 = 3.90
Henderson-Hasslebalch Equation: pH = pKa + log([salt]/[acid])
pKa = -log(6.3*10^(-5)) = 4.20
pH = 4.20 + log(0.10/0.20) = 4.20 - 0.30 = 3.90
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pH= pKa + log A-/HA and pKa=-log Ka
so -log 6.3x10^5 I am pretty sure should be 10^-5 is 4.20+(-.301) = 3.9 you would need to account for sig figs but that is close.
pH= pKa + log A-/HA and pKa=-log Ka
so -log 6.3x10^5 I am pretty sure should be 10^-5 is 4.20+(-.301) = 3.9 you would need to account for sig figs but that is close.
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