Hydrochloric acid and sodium hydroxide
HCl + NaOh ----> NaCl + H2O
The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5
The ph of the best buffer is 4.74
The given acetic acid is a weak acid
The equation of the pH of the buffer
pH = pKa + log ( conjugate base / weak acid ).
For best buffer the concentration of the weak acid and its conjugate base is equal.
pH = pKa + log 1
pH = pKa + 0
pH = pKa
given Ka = 1.8 × 10⁻⁵
pKa = - log ka
pH = -log ( 1.8 × 10⁻⁵ )
pH = 4. 74
Hence the pH of the best buffer is 4.74
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Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol
Molarity: mol/L
First, find mol.
62.5 g x 1mole ÷ 40.31 g = 1.55 mol
then divide mol and the given liters
1.55mol ÷ 1.50 L= 1.03 M
