solution number _______ volume of 0.10M NaCl used, mL _______ volume of 0.10M KC

Question

solution number _______ volume of 0.10M NaCl used, mL _______ volume of 0.10M KCl used, mL _______ [NaCl] _______ [KCl] _______ temperature of filtered KHT solution, degree C _______ molarity of standardized NaOH solution, mol L^-1 _______ volume of saturated KHT solution titrated, mL NaOH solution final buret reading, mL initial buret reading, mL volume of NaOH needed, mL number of moles of NaOH needed, mol number of moles of HT^- titrated, mol [HT^-]_sat, sol'n solubility of KHT, mol L^-1 [K^+]_initial [K^+]_from dissolved KHT [K^+]_sat sol'n K_sp average K_sp average solubility of KHT, mol L^-1 average solubility, grams per 100 mL

Explanation / Answer

Solution:

The computation are expressed below,

#1

No. of moles of NaOH = 0.10 M * 0.05 lt = 0.005 mol

# 2

No. of moles of NaOH = 0.10 M * 0.05 lt = 0.05 mol

# 1

No.of moles of HT- titrated = 16.5 / 74.551 = 0.2213 mol

# 2

No.of moles of HT- titrated = 17 / 74.551 = 0.228 mol

#1

[HT-] = 0.2213 mol / 1 lt = 0.2213 M

#2

[HT-] = 0.228 mol / 1 lt = 0.228 M

#1

Solubility of KHT = (0.10*0.2213)1/2 = 0.1487

#2

Solubility of KHT = (0.10*0.228)1/2 = 0.151

#1

[K] initial = 0.10 M

#2

[K] initial = 0.10 M

#1

[K] dissolved = 0.2213 M

#2

[K] dissolved = 0.228 M

#1

[K] salt = 0.3 M

#2

[K] salt = 0.3 M

#1

Ksp = 0.10*0.2213 = 0.002213

#2

Ksp = 0.10*0.228 = 0.00228

Average Ksp = (0.00213+0.228) / 2 = 0.002205

Average solubility KHT = (0.1487+0.151) / 2 = 0.150

Average solubility per 100 ml = 0.0150

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