A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sitting on a

Q: A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sitting on a

No of mol of buffer = 0.14*0.1 = 0.014 mol pH = pka + log(salt/acid) 5 = 4.76+log(x/(0.014-x)) x = 0.0088 mol No of mol of salt = 0.0088 mol No of mol of aceticacid = 0.014-0.0088 = 0.0052 mol after adding HCl . No of mol of HCl added = 5.4/1000*0.48 = 0.0026 mol pH = 4.76+log((0.0088-0.0026)/(0.0052+0.0026)) = 4.66 change in pH = 5-4.66 = 0.34

ID: 950000 • Letter: A

Question

A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.40 mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

The average human body contains 5.20 L of blood with a Fe2+ concentration of 2.00×105M . If a person ingests 8.00 mL of 17.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Explanation / Answer

No of mol of buffer = 0.14*0.1 = 0.014 mol

pH = pka + log(salt/acid)

5 = 4.76+log(x/(0.014-x))

x = 0.0088 mol

No of mol of salt = 0.0088 mol

No of mol of aceticacid = 0.014-0.0088 = 0.0052 mol

after adding HCl .

No of mol of HCl added = 5.4/1000*0.48

= 0.0026 mol

pH = 4.76+log((0.0088-0.0026)/(0.0052+0.0026))

= 4.66

change in pH = 5-4.66 = 0.34

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A beaker with 140 mL of an acetic acid buffer with a pH of 5.00 is sitting on a

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