I am confused about the molchange... where does the 0.004830 value come from? pH

Question

I am confused about the molchange... where does the 0.004830 value come from?

pH = - log [H+] = -log (9.7 times 10-3) = 2.01 The pH prior to the equivalence point (Weak Acid Buffer Problem) Between 0 mL and the equivalence point (54.17 mL in this example), the calculations performed are the same. These calculations are limiting reagent problems where the acid is in excess and the base is the limiting reagent. To illustrate this problem type, we will calculate the pH after 40.25 mL of NaOH have been added. In this particular case, this is a buffer problem because we have a weak acid (HF) and its (NaF).

Explanation / Answer

-0.004830 is the change due to addition of 0.004830 moles of NaOH which reacts with HF to give the corresponding salt

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