If 30.0 grams of solid NaF is placed in 1.00 liter of a 1.00 M HF solution, what is the pH of the buffer solution? For HA, Ka = 6.8x10-4

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anonymous · Answer

$$HF_{aq} + H_20_{(l)} = H_30_{(aq)}^++F_{(aq)}^-$$ We know that you are adding 30.0g of NaF and since Na is almost always a spectator we will disregard it and say you are adding 30g of F- to the HF solution. We first find moles of F-.$${30.0gF^-(mol) \over 18.998 g} = 1.6\ moles \ F^-$$ Then we use the Ka equation$$Ka = {[F^-][H_30^+] \over [HF]} $$$$6.8\times10^-4 = {[1.6 ][x] \over [1.0]} $$ x is your H+ concentration. You should be able to find pH from there. x was approximated to zero for the F- due to the fact that adding it would not have a reasonable effect.

anonymous · Answer

Thank you so much, I didn't know where to start.  I got the ph is 3.38.  I am going to post a questions that goes along with this, if you could help me get started.