A 40.3 ml sample of a 0.587 m aqueous hydrocyanic acid solution is titrated with a 0.363 m aqueous sodium hydroxide solution. The pH after 19.9 ml of base have been added is 12
After 19.9 mL of NaOH have been added we have gone way past the equivalence point, so there is a lot of excess NaOH,
⇒ 40.3 - 19.9 = 20.4 mL excess
⇒ 0.0204 L (0.363 M) = 0.0074 moles NaOH excess,
⇒ 0.0074 moles / (0.0199 + 0.0403)
⇒ 0.122 M NaOH at this point,
[OH]⁻ = [NaOH]
pOH = - log (0.122) = 2.103
pH = 14 - pOH = 11.897 = 12
Hence, pH = 12, After 19.9 mL base have been added.
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