I cannot figure out this problem, I've made a few attempts at solving it. I could...

I cannot figure out this problem, I've made a few attempts at solving it. I could really use some help. The question is : 75.00 mL of 0.225 M HNO2 is titrated to its equivalence point with 0.100 M NaOH. What is the pH at its' equivalence point? Please help!

Expert Solution

HNO2 + NaOH ---------------------> NaNO2 + H2O

75x 0.225 75x0.1 0 0

16.875 7.5 0 0 ------------------initial millimoles

9.375 0 7.5 7.5 -------------- at equivalent point

in the solution HNO2 + NaNO2 remained . so it is buffer

For acidic buffer

Henderson-Hasselbalch equation

pKa = -log Ka = -log(4.0 x 10-4) = 3.39

pH = pKa + log[salt/acid]

pH = 3.39 + log[7.5/9.375]

pH = 3.29

queen_honey_blossom answered 2 years ago

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