Given,
[tex]V = 5.2 L\\\\M = 1.1 KOH\\\\V_2 = 2.3 L\\\\M_2 = 0.2 Sc(OH)_3[/tex]
moles of each substance,
moles of base = [tex]M*V = 5.2L * 1.1M = 5.72 moles of KOH[/tex]
mole of [tex]Sc(OH)_3 = M*V = 2.3*0.2 = 0.46 moles of Sc(OH)_3[/tex]
Assume
[tex]Sc(OH)3 ---> Sc+3 and 3OH-[/tex]
therefore,
there is 3x the amount of moles of [tex]Sc(OH)_3[/tex] per mol of [tex]OH-[/tex]
[tex]0.46 * 3 = 1.38 mol of OH-[/tex]
Adding all OH- moles
[tex]1.38+5.72 = 7.1 mol of OH- present[/tex]
The total volume:
[tex]V_T = V_1+V_2 = 5.2+2.3 = 7.5 L[/tex]
The concentration of ions
[tex][OH-] = \frac{mol OH}{V}= \frac{7.1}{7.5} = 0.94667[/tex]
[tex]pOH = -log(OH-) = -log(0.94667) = 0.0238[/tex]
Therefore,
[tex]pH = 14 - pOH\\\\pH = 14 - 0.0238 = 13.97[/tex]
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