Respuesta :
for HClO, pKa = 7.54
for HNO_2, pKa = 3.15
for CH3 COOH, pKa = 4.74
Explanation:
The concentration of the solution given is 0.1 M has a pH closest to 7
The mixtures are weak acids and their salts except
HNO_3 and NaNO_3 pH = pH is near to '1'
for buffers( acidic) pH = pKa + log [salt] / [acid]
therefore [salt] = [acid] = 0.1 pH = pKa + log 0.1 / 0.1 = pKa
pH = pKa
for HClO, pKa = 7.54
for HNO_2, pKa = 3.15 therefore HClO and NaclO
mixture hs a pH closest to '7'
for CH3 COOH, pKa = 4.74
Answer:
The answer is: A. HCIO(aq) and NaCIO(aq
Explanation:
From buffers, the pH is equal:
pH = pKa + log([salt]/[acid])
[sal] = [acid] = 0.1 M
Replacing:
pH = pKa + log(0.1/0.1)
pH = pKa + 0
pH = pKa
According to this:
pKa of HClO ≅ 7.54
pKa of HNO₂ ≅ 3.15
pKa of CH₃COOH ≅ 4.74
Then, the mix that present a pH closest to 7 is HClO and NaClO