Calculate the p_H of a 1.2 times 10^-3 solution of H_2co_2, when 1.0 x 10^-2 NaH
ID: 501848 • Letter: C
Question
Calculate the p_H of a 1.2 times 10^-3 solution of H_2co_2, when 1.0 x 10^-2 NaHco_3, is added. K_01 of H_2CO_3, = 4.3 times 10^-7 16 Acid-Base Titrations A)strong acid-Strong Base calculate the p_oH and the pH of a solution in which 10.0 mL of 0.100 MHCI is added to 25.0 mL of 0.100 M NaOH. B) Weak Acid-Strong Base Calculate the [OH) and the p_H at the equivalence point for the titration of 500. mL of 0.10 M propionic acid (monoprotic) with 0.050 M calcium hydroxide K1 = 1.3 times 10^-5 17. Solubility Equilibria Mg(OH) (s)double arrow Mg^2 (aq) + 2 (OH) (aq)__Explanation / Answer
1) pH = pKa + log [NaHCO3]/[H2CO3]
pka = - log Ka = -log [4.3 x 10-7] = 6.37
pH = 6.37 + log [0.01] / [0.0012]
pH = 7.29
16) A) NaOH + HCl -----------> NaCl + H2O
millimoles of NaOH = 25 x 0.1 = 2.5
millimoles of HCl = 10 x 0.1 = 1.0
2.5 - 1.0 = 1.5 millimoles NaOH left
[NaOH] = 1.5 / 35 = 0.043 M
pOH = - log [OH-]
pOH = - log [0.043]
pOH = 1.37
pH = 14 - 1.37
pH = 12.63
B) millimoles of propionic acid = 500 x 0.1 = 50
50 millimoles of OH- must be added to reach equivalence point.
50 / 2 = 25 millimoles of Ca(OH)2 must be added
25 = Vx 0.05
V = 500 mL NaOH must be added
[salt] =50 / 1000 = 0.05 M
pH = 1/2 [pKw + pKa + log C]
pKa = - log Ka = - log [1.3 x 10-5] = 4.89
pH = 1/2 [14 + 4.89 + log 0.05]
pH = 8.79
pOH = 14 - 8.79
pOH = 5.21
[OH-] = 10-pOH = 10-5.21
[OH-] = 5.16 x 10-6 M
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