Calculate the molar solubility of Cr(OH)3 in 0.37 M NaOH; Kf for Cr(OH)4- is 8*1

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Explanation / Answer

Since, Cr(OH)3(s) ----> Cr+++(aq) + 3 OH-(aq) You need to get the value of KsP from the KSP table which is 6.3E-31 Ksp = 6.3E-31 = [Cr+++] * [OH-]^3 Now the Kf value is given for Cr(OH)4-, which can be derived from: Cr(OH)3(s) + OH-(aq) ----> Cr(OH)4-(aq) and now use the Kf value and calculate Kf = 8.0E+29 = [Cr(OH)4-] / ([Cr+++] * [OH-]^4) Kf * Ksp = 8.0E+29 * 6.3E-31 = 0.5 = [Cr(OH)4-] / [OH-] and now you get [Cr(OH)4-] = 0.5 * [OH-] since we will be using 0.15M NaOH solution here, therefore, [OH-] = 0.15 M So, [Cr(OH)4-] = 0.5 * 0.15 = 0.075 M

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