# How many mL (to the nearest mL) of 0.140-M KF solution should be added to 400. mL of 0.212-M HF to prepare a pH

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###### Question:

How many mL (to the nearest mL) of 0.140-M KF solution should be added to 400. mL of 0.212-M HF to prepare a pH

## Answers

Answer:205mL of 0.140M KF solutionExplanation:pH = 2.70 solution.It is possible to obtain the pH of the buffer of HF-KF using the H-H equation:pH = pKa + log [KF] / [HF]Where pH is desire pH = 2.70pKa is pKa of HF = 3.17[KF] could be taken as moles of KF And [HF] moles of HF: 400.0mL = 0.4L * (0.212mol/L) = 0.0848 moles of HFReplacing:2.70 = 3.17 + log [KF] / [0.0848 moles HF]-0.47 = log [KF] / [0.0848 moles HF]0.3388 = [KF] / [0.0848 moles HF][KF] = 0.02873 moles of KF must be added.In mL using concentration of KF (0.140M):0.02873 moles KF * (1L / 0.140 mol) = 0.205L =205mL of 0.140M KF solution

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