A beaker with 1.80×10 2 mL of an acetic acid buffer with a pH of 5.000 is sittin

Question

A beaker with 1.80×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.00 mL of a 0.340 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( ? ) sign if the pH has decreased.

PartA A beaker with 1.80x102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.00 mL of a 0.340 MHCl solution to the beaker. How much will he pH change? The pKa of acetic acid is 4.740 sign if the pH Express your answer numerically to two decimal places. Use a minus (- has decreased. View Available Hint(s) ApH Submit Previous Answers XIncorrect: Try Again; 2 attempts remaining Provide Feedback Next >

Explanation / Answer

As we know that;

By Using Hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

5 = 4.74 + log(conj.base/acid)

(conj.base)= 1.82(acid)

Then,

(acid) + (conj.base)= 0.1 M*1.80*10^2 ml = 18 mmol

(acid)+ 1.82(acid) = 18 mmol

(acid) = 18/2.82 = 6.4 mmol

(conj.base)= 18 - 6.4 = 11.6 mmol

Then,

when HCl = 0.360 M * 8.60 ml

= 2.924 m mol was added

pH = 4.74 + log((11.6 - 2.924)/(6.4 + 3.096))

= 4.71

pH change = initial - final

= 5 - 4.71= 0.29

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.