Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M HCHO_2 with 1

Question

Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M HCHO_2 with 100.0 mL of 0.20 M LiCHO_2. The K_a for HCHO_2 is 1.8 Times 10^-4. 10.13 3.74 3.47 3.87 10.53 Determine the molar solubility of AgBr in a solution containing 0.200 M NaBr. Ksp (AgBr) = 7.7 Times 10^-13. 0.200 M 8.8 Times 10^-7 M 3.8 Times 10^-12 M 1.54 Times 10^-13 M 5.8 Times 10^-5 M Determine the molar solubility of Agl in pure water. K_sp (Agl) = 8.51 Times 10^-17. 8.51 Times 10-17 M 2.77 Times 10^-6 M 4.40 Times 10^-6 M 4.26 Times 10^-17 M 9.22 Times 10^-9 M Determine the pH of a 0.00598 M HClO_4 solution. 7.566 6.434 3.558 2.223 11.777 Determine the pH of a 0.227 M C_5H_5N solution at 25 degree C. The K_b of C_5H_5N is 1.7 Times 10^-9. 4.71 9.29 10.14 9.41 4.59

Explanation / Answer

10.

0.227 M

Kb = 1.7 x 10-9

C5H5 is a weak base. In water it has equilibrium constant. The equilibrium is:

C5H5N(aq) + H2O(l) <=> C5H5NH+(aq) + OH-

Kb describes this equilibrium:

Kb = [C5H5NH+][OH-]/[C5H5N] = 1.7 x 10-9

For a monoprotic acid or base

Ka x Kb = Kw = 1.0 x 10-14

Let [OH-] = [C5H5NH+] = x, and let [C5H5N] = 0.227 – x;

Put these values into the equation for Kb.

(x)(x)/(0.227 - x) = 1.7 x 10-9

x2 = 1.7 x 10-9 (0.227 - x)

x2 = 1.7 x 10-9 (0.227)       

[Neglecting x term in the denominator since x << 0.227]

x2 = 1.7 x 10-9 (0.227)       

x2 = 3.86 x 10-10

x = 1.96 x 10-5

So,

x = 1.96 x 10-5 = [OH-]

[H+][OH-] = 1.0 x 10-14

[H+] = (1.0 x 10-14)/( 1.96 x 10-5) = 5.1 x 10-10 M

pH = -log(5.1 x 10-10) = 9.29

9.

Perchloric acid (HClO4) is a strong acid. It will dissociate completely.

HClO4     H+   + ClO4-

[H+] = 0.00598 M

pH = - log [H+] = - log (0.00598) =.2.22

8.

AgI(s)   Ag(aq) + I(aq)

note that [Ag] = [I] = x

Ksp = [Ag][I]

8.51 x 10-17 =x²

x = 9.22 x 10-9

[Ag] = 9.22 x 10-9

Solubility is therefore 9.22 x 10-9 mol/liter (quite insoluble).

6.

calculate the pH of a solution formed by mixing 250 mL of 0.15 M HCOOH with 100 mL of 0.2 M HCOOLi. Ka for HCOOH is 1.8 x 10-4

Molarity = moles/Liter

250 mL of 0.15 M HCOOH

Moles of HCOOH = 0.25 L x 0.15 M = 0.0375

100 mL of 0.2 M HCOOLi

Moles of HCOOLi = 0.10 L x 0.2 M = 0.0200

Total volume = 250 mL + 100 mL = 350 mL = 0.35 L

So,

Final concentrations are

[HCOOH] = 0.0375 mol / 0.35 L = 0 110 M

[HCOOLi] = 0.0200 mol / 0.35 L = 0 057 M

Ka = 1.8 x 10-4

pKa = -log Ka = - log (1.8 x 10-4) = 3.74

Using Henderson-Hesselbalch equation

pH = pKa + log [salt]/[acid]

pH = pKa + log [HCOOLi] / [HCOOH]

     = 3.74 + log (0 057 / 0 110)

     = 3.74 – 0.29

     = 3.45

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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