How many mL (to the nearest mL) of 0.203-M KF solution should be added to 580....

Question

Question

How many mL (to the nearest mL) of 0.203-M KF solution should be added to 580. mL of 0.202-M HF to prepare a pH = 3.70 solution?

Answer 1

Answer #1

No of moles of HF present in 580.0mL of 0.100 M of HF = 580.0 mL X- x 0.202 1000 mL = 0.117mol Ka of HF= 7.1x104 pKa=-logKa =

Answer 2

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