Chemistry, 14.11.2019 03:31 THEOUTCOMER

Abuffer is prepared by adding 150 ml of 1.0 m naoh to 250 ml of 1.0 m nah2po4. how many moles of hcl must be added to this buffer solution to change the ph by 0.18 units? assume the total volume remains unchanged at 400 ml. for h2po4-, ka2 = 6.3 × 10^-8.
o 0.063 mol
o 1.0 mol
o 0.025 mol
o 0.082 mol
o 0.50 mol

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Answers

Answer from: jmallory3031

0.025 mol

Explanation:

When the NaH₂PO₄ reacts with NaOH, first it will dissociate in Na⁺ and H₂PO₄⁻, in a 1:1:1 reaction

NaH₂PO₄ → Na⁺ + H₂PO₄⁻

The number of moles of H₂PO₄⁻ is the number of moles of NaH₂PO₄

n = 0.250 L * 1 mol/L = 0.250 mol

The number of moles of NaOH that reacts is:

n = 0.150 L * 1 mol/L = 0.150 mol

The base will react with the acid H₂PO₄⁻ and will form the anion HPO₄²⁻ which is the conjugate base of the acid. The acid and it's conjugate base will form the buffer.

Let's suppose that all the base reacts, so, the number of moles of the substances of interest will be:

nH₂PO₄⁻ = 0.250 - 0.150 = 0.100 mol

nHPO₄²⁻ = 0.150 mol

The pH of a buffer can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[A⁻]/[HA]

Where pKa = -logKa, [A⁻] is the conjugate base concentration (HPO₄²), and [HA] is the acid concentration (H₂PO₄⁻)

If the pH change for 0.18 units, the concentrations of A⁻ and HA will change too. So, let's call them [A⁻]f and [HA]f. Because an acid is being added, the pH must decrease, so pH final - pH initial = -0.18

(pKa + log[A⁻]f/[HA]f) - (pKa + log[A⁻]/[HA]) = -0.18

The volume doesn't change, so we can replace the concentration by the number of moles. The acid will react with the A⁻ and will form HA so:

A⁻f = 0.150 - nHCl

HAf = = 0.100 + nHCl

pKa - log((0.150 - nHCl)/(0.100 + nHCl)) - pKa - log(0.150/0.100) = -0.18

log((0.150 - nHCl)/(0.100 + nHCl)) - log(0.150/0.100) = - 0.18

log((0.150 - nHCl)/(0.100 + nHCl))/(0.150/0.100)= -0.18

((0.150 - nHCl)/(0.100 + nHCl))/(0.150/0.100)= $10^{-0.18}$