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Consider a buffer made by adding 132.8 g of nac₇h₅o₂ to 300.0 ml of 2.31 m hc₇h₅o₂ (ka = 6.3 x 10⁻⁵) what is the ph of the buffer after 0.750 mol of h⁺ have been added?

Sagot :

The ph of the buffer solution given after 0.750 mol of H⁺ have been added is; 4.34

What is the PH of the Buffer?

pKa = -log(Ka) = 4.20

[A⁻] = [NaC₇H₅O₂]

[HA] = [HC₇H₅O₂]

Concentration of NaC₇H₅O₂ in the preferred buffer solution is;

[132.8g/(144.11 g/mol * 300 mol)] * 1000 mL/1 L = 3.07 M

Formula for the PH here is;

pH = pKa + log [A⁻/HA]

Thus;

pH = -log(6.3 x 10⁻⁵) + log [3.07/2.31]

pH = 4.34

Read more about bufffer PH at; https://brainly.com/question/22390063

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