Data Sheet solution number H (e volume of 0.10M Naci used, mL volume of 0.10M KC

Question

Data Sheet solution number H (e volume of 0.10M Naci used, mL volume of 0.10M KCl used, mL 50m KHT (NaCij 2.003 [KCll temperature of filter KHT solution, C molarity of standardized NaoH solution, mol L 1 08 13 determination volume of saturated KHT solution titrated, mL 2 NaOH solution to OO final buret reading, mL initial buret reading, mL 8.00 roo volume of NaOH needed, mL number of moles of NaOH needed, mol number of moles of HT titrated, mol [HT lsat. sorn solubility of KHT, mol L-1 MK linitial from dissolved KHT [K lsat sorn average Ksp average solubility of KHT, mol L average solubility, grams per 100 mL

Explanation / Answer

1) determination NO 1.

Number of moles of NaOH = MV = 0.0813*8 = 0.6504 mmol of NaOH = 0.6504*10^-3 mol of NAOH

moles of HT titrated --> if atio is 1:1 then 0.6504 mmol of HT titrated = 0.6504*10^-3 mol

[Ht-]] sat solution = mol/V = (0.6504*10^-3) / 36 = 0.00001806666 M of HT-

solubility --> 0.00001806666 mol per liter

[K+] = initial --> 0.05 M (from [KCl] on top)

[K+] from KHT --> + 0.00001806666

[K+] initial --> 0.05

[K+] sat. sonl = 0.05 + 0.00001806666 = 0.05002

Ksp = S^2 = (0.00001806666 )(0.00001806666 ) = 3.264*10^-10

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