When a 25. 0-ml sample of 0. 150 M hydrofluoric acid is titrated with a 0. 150 M naoh solution,the ph after 13. 3 ml of base is added 1.33.
As we already know that sodium hydroxide, which we will refer here to as [tex]OH^{-}[/tex] since it is a strong base, and hydrofluoric acid, HF, a weak acid, react in a mole ratio of 1:1 to form fluoride anions.
also, using molarity , we see
moles of HF = [tex]25*\frac{0.150}{10^{3} }[/tex] = 0.00375
moles of [tex]OH^{-}[/tex] = [tex]13.3*\frac{0.150}{10^{3} }[/tex] = 0.001995
Observe that there are now more moles of hydrofluoric acid. In this case, hydroxide anions will operate as the limiting reagent.
The finished reaction will produce a solution that contains;
0.001995moles - 0.001995moles = 0 moles [tex]OH^{-}[/tex]
0.00375moles - 0.001995moles = 0.001755 moles HF
There will be no remaining hydroxide anions
The final solution will have a total volume of
25 mL + 13.3mL = 38.3mL
This means that the concentration of the hydrofluoric acid in the resulting solution will be,
[HF] = [tex]\frac{0.001755}{38.3*10^{-3} }[/tex]
= 0.0458mol/l
pH = -log [[tex]H^{+}[/tex]]
pH = -log [ 0.0458]
pH = 1.33
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