If 0.81g of NaClO2 is added to 150 mL of 0.11M HClO2, what is the pH of the buffer? ( Write answer to the hundredths place) KA(HClO2) = 1.1x10-2
Molar mass of NaClO2,
MM = 1*MM(Na) + 1*MM(Cl) + 2*MM(O)
= 1*22.99 + 1*35.45 + 2*16.0
= 90.44 g/mol
mass(NaClO2)= 0.81 g
use:
number of mol of NaClO2,
n = mass of NaClO2/molar mass of NaClO2
=(0.81 g)/(90.44 g/mol)
= 8.956*10^-3 mol
volume , V = 1.5*10^2 mL
= 0.15 L
use:
Molarity,
M = number of mol / volume in L
= 8.956*10^-3/0.15
= 5.971*10^-2 M
Ka = 1.1*10^-2
pKa = - log (Ka)
= - log(1.1*10^-2)
= 1.959
use:
pH = pKa + log {[conjugate base]/[acid]}
= 1.959+ log {5.971*10^-2/0.11}
= 1.693
Answer: 1.69
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