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.)

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× = 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× = 2x10⁻⁴ L of HA

2x10⁻⁴ L of HA× = 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!

the correct answer is:

a. formation of precipitation

the answer is d. 74.02