Calculate the p-value for each of the ions in the following: A. Na+, Cl-, and OH

Question

Calculate the p-value for each of the ions in the following: A. Na+, Cl-, and OH- in a solution that is 0.0635M in NaCL and 0.0403 M in NaOH. B. Ba^2+, Mn^2+, and Cl^- in a solution that is 4.65* 10^-3M in BaCl2 and 2.54 M in MnCl2. Calculate the p-value for each of the ions in the following: A. Na+, Cl-, and OH- in a solution that is 0.0635M in NaCL and 0.0403 M in NaOH. B. Ba^2+, Mn^2+, and Cl^- in a solution that is 4.65* 10^-3M in BaCl2 and 2.54 M in MnCl2. A. Na+, Cl-, and OH- in a solution that is 0.0635M in NaCL and 0.0403 M in NaOH. B. Ba^2+, Mn^2+, and Cl^- in a solution that is 4.65* 10^-3M in BaCl2 and 2.54 M in MnCl2.

Explanation / Answer

M =0.0635 of Na+ and Cl-

M = 0.0403 M of Na+ and OH-

Total [Na+] = 0.0635 +0.0403 = 0.1038

Total [Cl-] = 0.0635

Total [OH-] = 0.0403

pNa = -log(Na+) = -log(0.1038) = 0.983

pCl- =  -log(Cl-) = -log(0.0635) = 1.1972

pOH =  -log(OH-) = -log(0.0403 ) = 1.39469

For 2)

[Ba+2] = 4.65*10^-3

[Mn+2] = 2.54

[Cl-] = 2*( 4.65* 10^-3 + 2*(2.54) = 10.1693

pBa+2 = -log(Ba+2) = -log(4.65*10^-3) = 2.3325

pMn+2 = -log(Mn+2) = -log( 2.54) = -0.40483

pCl- =  -log(Cl-) = -log(10.1693) = -1.007291

NOTE: those are negative since they are too high, the log scale is based on 10^p

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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