Calculate [OH-] in a solution of: a) sodium hydroxide with pH of 8 b) calcium hy

Question

Calculate [OH-] in a solution of: a) sodium hydroxide with pH of 8 b) calcium hydroxide with pH of 12 c) hydrochloric acid with pH of 5 Show work please!!!

Explanation / Answer

Calculate the pH of a 1 x 10^-8 M NaOH(aq) solution We must consider the water dissociation H2O H+ + OH- [H+]= [OH-] Kw = [H+][OH-]= 1.0 x 10^-14 [OH-]= 1.0 x 10^-7 b Calculate the molar solubility of calcium hydroxide in a solution buffered Calcium hydroxide dissolves and dissociates according to: Ca(OH)2(s) ? Ca²?(aq) + 2 OH?(aq) So the ionic molarities in a saturated solution satisfy the following equilibrium equation Ksp = [Ca²?] · [OH?]² The solubility constant at 25°C is [1] Ksp = 5.02×10^–6 Let s be the molar solubility of Ca(OH)2. When you dissolve s moles per liter, all the salt dissolves and dissociates, forming one calcium ion per salt molecule. Hence the calcium ion molarities is: [Ca²?] = s The hydroxide ion concentration does not change due to buffering. It is fixed to a level, which is specified by the given pH: [OH?] = 10^(pH - 14) Therefore Ksp = s · (10^(pH - 14))² = s · 10^(2·pH - 28) => s = Ksp / 10^(2·pH - 28) = Ksp · 10^(28 - 2·pH) at pH=5 s = 5.02×10^–6 · 10^(28 - 2·5) = 5.02×10^–6 · 10^(18) = 5.02×10^12 mol/L at pH=7 s = 5.02×10^–6 · 10^(28 - 2·7) = 5.02×10^–6 · 10^(14) = 5.02×10^8 mol/L at pH=8 s = 5.02×10^–6 · 10^(28 - 2·7) = 5.02×10^–6 · 10^(12) = 5.02×10^6 mol/L PH of .01M of Ca(OH)2? Calcium hydroxide is indeed a strong base, which means that every bit of it that dissolves in water dissociates into calcium ions and hydroxide ions. BUT, calcium hydroxide isn't very soluble in water. Therefore we can use the Ksp of calcium hydroxide to find the concentration of OH- in a saturated solution. Ksp Ca(OH)2 = 4.7 x 10^-6 Ksp = [Ca2+] [OH-]^2 = (x) (2x)^2 = 4.7 x 10^-6 4x^3 = 4.7 x 10^-6 x = 0.0106 Therefore a saturated solution has an [OH-] of 0.0212 M Based on the concentration of Ca(OH)2, we can assume that the hydroxide ion concentration is 0.020 M (twice 0.010 M) which is very close to being saturated. A hydroxide ion concentration of 0.0200 gives a pH of 12.30. NaOH is a strong base NaOH => Na+ + OH- [OH-] = 1 x 10^-8 M total concentration OH- = 1 x 10^-7 + 1 x 10^-8 =1.1 x 10^-7 M pOH = 6.96 pH = 14 - 6.96 =7.04

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