Answer to Question #170044 in General Chemistry for Mae

Question #170044

A 555 mL of 0.500 M lactate buffer (Ka = 1.38 x 104), pH 4.0 is to be prepared. How many moles of NaOH is needed to prepare the buffer?


1
Expert's answer
2021-03-09T06:04:08-0500

Ka = 1.38×10-4

pKa =100.139 × 10-4 = 4-0.139 = 3.861

pH = pKa + log"\\frac{salt}{acid}"


4 = 3.861 + log"\\frac{salt}{0.555\u00d70.5}"


1.38 = "\\frac{salt}{0.2775}"


moles of salt = 1.38×0.2775 = 0.38295M


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