There is a solution of 6.2×10−2 M of KOH?
1. Determine [H3O+]
2. Determine pH
3. Determine products when reacted with excess of H3PO4-
4.Determine milliliters required to neutralize 44.0mL of 6.0×10−2 H2SO4 (in mL)
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[H3O+] = 1.0 x 10^-14 / 6.2 x 10^-2 = 1.6 x 10^-13
pH = - log 1.6 x 10^-13 = 12.8
K3PO4
Moles H2SO4 = 0.0440 x 6.0 x 10^-2 = 0.00264
moles KOH needed = 2 x 0.00264 = 0.00528
V = 0.00528 / 6.2 x 10^-2 = 0.085 L => 85 mL