Answer:
pAl³⁺ = 1,699
pPb²⁺ = 1,866
Explanation:
In this problem, the first titration with EDTA gives the moles of Al³⁺ and Pb²⁺, these moles are:
Al³⁺ + Pb²⁺ in 25,00mL = 0,04907M×0,01714L = <em>8,411x10⁻⁴ moles of EDTA≡ moles of Al³⁺ + Pb²⁺.</em>
The molar concentration is <em>8,411x10⁻⁴ moles/0,02500L = </em><em>0,0336M Al³⁺+Pb²⁺.</em>
In the second part each Al³⁺ reacts with F⁻ to form AlF₃. Thus, you will have in solution just Pb²⁺.
The moles added of EDTA are:
0,02500L×0,04907M = 1,227x10⁻³ moles of EDTA
The moles of EDTA in excess that react with Mn²⁺ are:
0,02064M × 0,0265L = 5,470x10⁻⁴ moles of Mn²⁺≡ moles of EDTA
That means that moles of EDTA that reacted with Pb²⁺ are:
1,227x10⁻³ moles - 5,470x10⁻⁴ moles = 6,800x10⁻⁴ moles of EDTA ≡ moles of Pb²⁺.
The molar concentration of Pb²⁺ is:
6,800x10⁻⁴mol/0,0500L = <em>0,0136 M Pb²⁺</em>
Thus, molar concentration of Al³⁺ is:
0,0336M Al³⁺+Pb²⁺ - 0,0136 M Pb²⁺ = <em>0,0200M Al³⁺</em>
pM is -log[M], thus pAl³⁺ and pPb²⁺ are:
<em>pAl³⁺ = 1,699</em>
<em>pPb²⁺ = 1,866</em>
I hope it helps!
