when 15.12 ml of 0.1194 M NaOH is added to 25 ml of 0.1264 M HX(an unnknown weak

Question

when 15.12 ml of 0.1194 M NaOH is added to 25 ml of 0.1264 M HX(an unnknown weak acid),the resulting pH is 5.9 . what is the ka of the unknown acid?

Explanation / Answer

First calculate moles of NaOH and moles of HX. moles NaOH = M NaOH x L NaOH = (0.1194)(0.01512) = 0.001805 moles NaOH. moles HX = M HX x L HX = (0.1246)(0.02500) = 0.003115 moles HX. HX and NaOH react as follows: HX + NaOH ==> H2O + NaX Since the HX and NaOH react in a 1:1 mole ratio, then the 0.003115 moles of HX will completely use up the 0.001805 moles of NaOH. You will have (0.003115 - 0.001805) = 0.001310 moles of HX left and you will have formed 0.001805 moles of NaX. The combination of HX and NaX is a buffer system. The pH of this buffer is given by the Henderson-Hasselbalch equation: pH = pKa + log (moles NaX / moles HX) 5.9 = pKa + log (0.001805 / 0.001310) 5.9 = pKa + 0.139 5.76 = pKa Ka = 10^-5.76 = 1.7 x 10^-6

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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