Question

-2 you mix 125 ml o milliliters of 1.3M HCl would have to be added to lower the pH of this mixture by .3?

Answer 1

Answer:

mmol of HCO₃^- = Molarity*volume in mL = 3.0*125.0 = 375.0 mmol

mmol of CO₃^2- = 3.0*145.0 = 435.0 mmol

pKa (HCO₃^- ) = pKa2 (H₂CO₃) = 10.33

pH of the buffer is given by Henderson-Hasselbalch equation as,

pH = pKa + log(mmol of CO₃^2-/ mmol of HCO₃^-)

pH = 10.33 + log(435 / 375)

pH = 10.33 + 0.06

pH = 10.39

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We need to lower this pH by 0.3 unit

So Expected pH = 10.39 - 0.3 = 10.09

Let A mmol of HCl added to above buffer to lower pH to 10.09

Addition of HCl will increase # of mmol of HCO₃^-.by "A" mmols and decrease # of mmol of CO₃^2-? by "A" mmols.

mmol of HCO₃^- = 375.0+A mmol of CO₃^2- = 435-A.

Using Henderson's equation

pH = pKa + log(mmol of CO₃^2-/ mmol of HCO₃^-).

10.09 = 10.33 + log (435-A / 375+A)

.log (435-A / 375+A) = -0.22

(435-A / 375+A)? = 10^-0.22.

(435-A / 375+A)? = 0.603

By compnendo

810.0 / 375+A = 1.603

1.603(375+A) 810.0

1.603A = 810.0 - 1.603*375.

1.603A = 208.875

A = 208.875 / 1.603

A = .130.3 mmol

Now,

mmol of HCl = Molarity of HCl*Volume of HCl in mL

Volume of HCl i mL mmol of HCl / Molarity = 130.3 / 1.3 = 100.2 mL

100.2 mL HCl eeded to be added

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