A25.00 ml sample containing an unknown amount of al3+ and pb2+ required 17.14 ml of 0.04907 m edta to reach the end point. a 50.00 ml sample of the unknown was then treated with f− to mask the al3+ . to the 50.00 ml sample, 25.00 ml of 0.04907 m edta was added. the excess edta was then titrated with 0.02064 m mn2+ . a total of 26.5 ml was required to reach the methylthymol blue end point. determine pal3+ and ppb2+ in the unknown sample.

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 = 8,411x10⁻⁴ moles of EDTA≡ moles of Al³⁺ + Pb²⁺.

The molar concentration is 8,411x10⁻⁴ moles/0,02500L = 0,0336M Al³⁺+Pb²⁺.

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 = 0,0136 M Pb²⁺

Thus, molar concentration of Al³⁺ is:

0,0336M Al³⁺+Pb²⁺ - 0,0136 M Pb²⁺ = 0,0200M Al³⁺

pM is -log[M], thus pAl³⁺ and pPb²⁺ are:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

I hope it helps!

heat or electricity in the atmosphere- c.