# A 1.50 L buffer solution is 0.250 M HF and 0.250 M NaF. What is the pH of the solution after the addition of 0.0500 moles of solid NaOH? Assume no volume change. Ka for HF is 3.5×10^-4

##”pH” = 3.57##

Your buffer contains hydrofluoric acid, ##”HF”##, weak acid, and sodium fluoride, ##”NaF”##, the salt of its conjugate base, the fluoride anion, ##”F”^(-)##.

When the sodium hydroxide solution is added, assuming with no change in the total volume of the buffer, you can expect the weak acid and the strong base to each other.

Moreover, this reaction will result in the production of more conjugate base.

So, the balanced chemical equation for the reaction between hydrofluoric acid and sodium hydroxide looks like this

##”HF”_text((aq]) + “NaOH”_text((aq]) -> “NaF”_text((aq]) + “H”_2″O”_text((l])##

Notice that ##1## mole of hydrofluoric acid will react with ##1## mole of sodium hydroxide and produce ##1## mole of sodium fluoride.

Use the of the hydrofluoric acid and the volume of the buffer to determine how many moles you have in solution

##color(blue)(c = n/V implies n = c * V)##

##n_(HF) = “0.250 M” * “1.5 L” = “0.375 moles HF”##

Do the same for the conjugate base

##n_(F^(-)) = “0.250 M” * “1.5 L” = “0.375 moles F”^(-)##

Now, you’re adding ##0.0500## moles of sodium hydroxide to the buffer. Since you have fewer moles of strong base than on weak acid, it follows that the sodium hydroxide will be completely consumed.

The number of moles of hydrofluoric acid will be changed to

##n_(HF) = “0.375 moles” = “0.0500 moles” = “0.325 moles”##

The number of moles of fluoride anions will increase by the same amount

##n_(F^(-)) = “0.375 moles” + “0.0500 moles” = “0.425 moles F”^(-)##

Use the volume of the buffer to calculate the new molarities of the weak acid and its conjugate base

##[“HF”] = “0.325 moles”https://studydaddy.com/”1.5 L” = “0.21667 M”##

##[“F”^(-)] = “0.425 moles”https://studydaddy.com/”1.5 L” = “0.28333 M”##

Finally, use the Henderson – Hasselbalch equation to find the of the buffer

##color(blue)(“pH” = pK_a + log( ([“conjugate base”])/([“weak acid”]))##

Use the acid dissociation constant, ##K_a##, to get the value of ##pK_a##

##pK_a = – log(K_a)##

The pH of the solution will thus be

##”pH” = -log(K_a) + log( ([“F”^(-)])/([“HF”]))##

##”pH” = -log(3.5 * 10^(-4)) + log( (0.28333color(red)(cancel(color(black)(“M”))))/(0.21667color(red)(cancel(color(black)(“M”)))))##

##”pH” = color(green)(3.57)##

Finally, does this result make sense?

Notice that your starting buffer had equal concentrations of weak acid and conjugate base. This means that the H-H equation can be reduced to

##”pH” = pK_a + log(1) = pK_a##

Initially, the pH of the solution was equal to ##3.46##.

The buffer then converts the strong base to weak base, and so the concentration of conjugate base is bigger than that of the weak acid. This is why the pH of the buffer increases after the addition of the strong base.

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