Answer:
The moles of A⁻ present in the mixture are 3,20x10⁻⁵ in a format 3,20E-5
Explanation:
A buffer is a mixture of a weak acid (In this case, HA) with its conjugate base.
The moles of weak acid are:
5,00 mL× = 5x10⁻³ L of HA
5x10⁻³ L of HA× = 5x10⁻⁵ mol of HA
The initial moles of conjugate base are:
3,00 mL× = 3x10⁻³ L of NaA
3x10⁻³ L of HA× = <em>3x10⁻⁵ mol of NaA</em>
The reaction of a weak acid with a strong base as NaOH produce:
HA + NaOH → NaA + H₂O
0.20 mL 0.010 M NaOH are:
0,20 mL× = 2x10⁻⁴ L of HA
2x10⁻⁴ L of HA× = <em>2x10⁻⁶ mol of NaOH</em>
Each mol of NaOH is producing NaA. That means:
<em>2x10⁻⁶ mol of NaOH ≡ 2x10⁻⁶ mol of NaA</em>
Thus, moles of A⁻ present in the mixture are:
<em>2x10⁻⁶ mol of NaA + 3x10⁻⁵ mol of NaA = </em><em>3,20x10⁻⁵ mol of NaA, in E-format: 3,20E-5</em>
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I hope it helps!