Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.1 m in kh2po4, what concentration of na2hpo4 would you need? (pka for h3po4, h2po−4, and hpo2−4 are 2.14, 6.86, and 12.4, respectively.)

0.14 M

Explanation:

A buffer is a solution where acid is in equilibrium with its conjugate base (or a base with its conjugate acid), and because of that, when acids or bases are added to it, the pH remains almost unaltered.

In this situation, the salts mixed will dissociate, and form the ions H₂PO₄⁻ and HPO₄²⁻. The original acid is the polyprotic acid H₃PO₄, which has 3 dissociations:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺

H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺

HPO₄²⁻ ⇄ PO₄³⁻ + H⁺

So, in this case, the second equilibrium is presented, H₂PO₄⁻ is the acid and HPO₄²⁻ is the conjugate base, with pKa = 6.86. The value of the pH of a buffer can be determined by the Handerson-Halsebach equation:

pH = pKa + log[A-]/[HA]

Where [A-] is the concentrantion of the conjugate base, and [HA] is the concentrantion of the acid. The salts have stoichiometry 1:1 with the ions given, so the concentration of the salt is the concnetration of the ion, [HA] = [H₂PO₄⁻] = [KH₂PO₄] = 0.1 M:

7.00 = 6.86 + log[A-]/0.1

log[A-]/0.1 = 0.14

[A-]/0.1 =

[A-]/0.1 = 1.38

[A-] = 0.14 M

[Na₂HPO₄] = 0.14 M

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2