Question

Calculate the h + ion concentration in 0.1 molar acetic acid solution given that the dissociation constant of acetic acid in water is 1.8 into 10 to the power minus 5

Answer 1

pKa = -log(10)[Ka]

Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

as [H+] = [CH3COO-]

pKa = log(10)[CH3COOH] - 2*log(10)[H+]

For acetic acid pKa = 4.76 and -log(10)[H+] = pH

4.76 = log(10)(0.1) + 2*pH

=> 2*pH = 4.76 + 1

=> pH = 5.76/2

= 2.88

Answer 2

pKa = -log(10)[Ka]

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]pKa = log(10)[CH3COOH] - 2*log(10)[H+]

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]pKa = log(10)[CH3COOH] - 2*log(10)[H+]For acetic acid pKa = 4.76 and -log(10)[H+] = pH

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]pKa = log(10)[CH3COOH] - 2*log(10)[H+]For acetic acid pKa = 4.76 and -log(10)[H+] = pH4.76 = log(10)(0.1) + 2*pH

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]pKa = log(10)[CH3COOH] - 2*log(10)[H+]For acetic acid pKa = 4.76 and -log(10)[H+] = pH4.76 = log(10)(0.1) + 2*pH=> 2*pH = 4.76 + 1

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]pKa = log(10)[CH3COOH] - 2*log(10)[H+]For acetic acid pKa = 4.76 and -log(10)[H+] = pH4.76 = log(10)(0.1) + 2*pH=> 2*pH = 4.76 + 1=> pH = 5.76/2

pKa = -log(10)[Ka]Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]as [H+] = [CH3COO-]pKa = log(10)[CH3COOH] - 2*log(10)[H+]For acetic acid pKa = 4.76 and -log(10)[H+] = pH4.76 = log(10)(0.1) + 2*pH=> 2*pH = 4.76 + 1=> pH = 5.76/2= 2.88