2. How much water, in liters, must be added to 0.50 liter of 6.0 M HCl to make the solution 2.0 M
1 answer:
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Answer:
a) 5,3176x10⁻⁴ moles
b) 6,85x10⁻⁴ moles
c) The appropriate formula to calculate is Henderson-Hasselbalch.
d) pH = 4,86. Acidic solution but slighty
Explanation:
a) moles of acetic acid:
9,20x10⁻³L × 57,8x10⁻³M = <em>5,3176x10⁻⁴ moles</em>
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b) moles of sodium acetate:
56,2x10⁻³g ÷ 82,0 g/mole = <em>6,85x10⁻⁴ moles</em>
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c) The appropriate formula to calculate is Henderson-Hasselbalch:
pH= pka + log₁₀
d) pH= 4,75 + log₁₀
<em>pH = 4,86</em>
<em>3 < pH < 7→ Acidic solution but slighty</em>
I hope it helps!