Calculate the pH of 3.8 times 10^-13 M OH^-. 12.41 15.59 2.06 7.00 1.58 A 25.0 m

Question

Calculate the pH of 3.8 times 10^-13 M OH^-. 12.41 15.59 2.06 7.00 1.58 A 25.0 mL sample of H_2SO_4(sq) is titrated with 15.5 mL of 0.50 M NaOH(sq). what is the concentration of the H_2SO_4(sq)? 0.078 M 0.16 M 0.31 M 0.62 M A buffer solution contains carbonic acid (H_2CO_3) at a concentration of0.0050 M and sodium bicarbonate (NaHCO_3) at a concentration of 0.10 M. The relevant equilibrium is shown below. What is the pH of this buffer solution? H_2CO_3(aq) + H_2O(l) ReverseEquilibrium H_3O^+(aq) + HCO_3^-(aq) k_a = 4.5 times 10^-7 4.50 7.00 6.35 7.65 2.16 Extra credit: Write the reaction of HSO_4^- (weak base) with water on the back of your scantron.

Explanation / Answer

18.

[OH-] = 3.8*10^-13

then

pOH = -log([OH-])

pOH = -log(3.8*10^-13) = 12.4202

pH = 14-12.4202 = 1.58

then

pH = 1.58

NOTE: please consider posting multiple questions in multiple set of Q&A. We are not allowed to answer to multiple questions in a single set of Q&A.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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