Moles of H⁺ released by each mole of acid = 3
Moles of H⁺ released = 3
Moles of OH⁻ released = 1.75
Moles of H⁺ remaining = 3 - 1.75 = 1.25 mol/dm³
pH = -log[H⁺]
pH = -log(1.25)
pH = -0.1
Did you mean significant figures? If so, 67.6(1.2)=81
When multiplying, you always leave the the answer in the least amount of significant figures.
Answer:
(−ΔH°1)+12(−ΔH°2)+(ΔH°3)+(ΔH°4)
I hope this helps you out!! :)
