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

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

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3.7 Ratings (716 Votes)

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

75x 0.225Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 75x0.1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

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

9.375Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 7.5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 7.5Â Â -------------- 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

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