Chemistry Acid/ Base HW Question?

A 25.0-mL test of 0.10 M C2H3NH2 (ethylamine) is titrated with 0.15 M HCl. What’s the pH for the

Perform some stoichiometry initially: know which compound is within extra (I prefer B to portray the beds base C2H3NH2, BH+ to portray its conjugate acid C2H3NH3+) —
HCl + B –> BH+ + Cl-
Moles Acid HCl = 0.15 mol/L x .00900L = 0.00135 mol
Moles Base B = 0.10 mol/L x .0250L = 0.0025 mol
Acid are used-up; 0.00135 mol BH+ are created.(1 to at least one mole proportion)
Base B is within extra: .00115 mol.
Today utilize the balance appearance the dissociation for the poor base:
Kb = [OH-][BH+] / [B] = 6.5 x 10-4
[B] = moles B / brand new amount = .00115mol/.0340L = .0338mol/L
[BH+] = moles BH+/new amount = .00135mol/.034L = .0397mol/L
[OH-] = x (unknown)
Kb = [.0397][x]/[.0338]
x = (6.5×10-4)[.0338]/[.0397]
x = [OH] = [.00055]
pOH = 3.26, pH = 14 – pOH = 10.74
Training this! These issues are difficult to start with, but get simpler should you choose a number of consecutively. STOICH initially, after that balance…

C. i acquired equivalent concern in chemisty….

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