You have prepared a buffer by combining 5.00 mL 0.010 M HA (weak acid) and 3.00 mL 0.010 M NaA (provides A -, the conjugate base of HA). pK a for HA is 6.23. How many moles of A - will be present in the mixture after addition of 0.20 mL 0.010 M NaOH? (Suggestion: Start by calculating the moles of everything in the mixture.) Enter the numerical value, without units, to at least 2 significant figures. You may use E-format scientific notation, e.g., 1.23x10 -4 would be entered as 1.23E-4. (Note that the acid and base do not add up to 20.00 mL -- water must be added. That does not affect how you answer the question.)

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jufsanabriasa jufsanabriasa · Answer 1 · 2019-10-24T13:33:52+02:00

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×[tex]\frac{1L}{1000mL}[/tex] = 5x10⁻³ L of HA

5x10⁻³ L of HA×[tex]\frac{0,010mol}{1L}[/tex] = 5x10⁻⁵ mol of HA

The initial moles of conjugate base are:

3,00 mL×[tex]\frac{1L}{1000mL}[/tex] = 3x10⁻³ L of NaA

3x10⁻³ L of HA×[tex]\frac{0,010mol}{1L}[/tex] = 3x10⁻⁵ mol of NaA

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×[tex]\frac{1L}{1000mL}[/tex] = 2x10⁻⁴ L of HA

2x10⁻⁴ L of HA×[tex]\frac{0,010mol}{1L}[/tex] = 2x10⁻⁶ mol of NaOH

Each mol of NaOH is producing NaA. That means:

2x10⁻⁶ mol of NaOH ≡ 2x10⁻⁶ mol of NaA

Thus, moles of A⁻ present in the mixture are:

2x10⁻⁶ mol of NaA + 3x10⁻⁵ mol of NaA = 3,20x10⁻⁵ mol of NaA, in E-format: 3,20E-5

I hope it helps!