Ye M 2 24.60 mL of 0.185 M HNO, are titrated with 27.35 mL of a KOH solution. Wh
ID: 1039142 • Letter: Y
Question
Ye M 2 24.60 mL of 0.185 M HNO, are titrated with 27.35 mL of a KOH solution. What is the molari KOH? 2?·35 2. 0.280 gram of KOH will just neutralize what volume of 0.200 M H2SO4? 3.0.750 gram of a sample of commercial lye is dissolved in water and titrated with 32.00 mL 0.500 M HCI. What is the percent purity (the active ingredient in Lye is NaOH) of this sample 32.oo ml 4. A 0.345 gram sample of oxalic acid dihydrate crystals (H C:0 2 HO) is dissolved in w titrated with 24.50 mL of a sodium hydroxide solution. What is the molarity of the NaOH? AAM uzczou. ??10 12 80.0 mL of 0.200 M NaOH is mixed with 20.0 mL of 0.600 M HCL. What is the conce maining OH? (Final volume is 100.0 mL.) NExplanation / Answer
2)
moles = mass / molar mass
= 0.280 / 56.1
= 4.99 x 10^-3
2 KOH + H2SO4 -----------------------> K2SO4 + 2H2O
2 mole ------------------> 1 mole H2SO4
4.99 x 10^-3 mole KOH = 4.99 x 10^-3 / 2 = 2.50 x 10^-3
molarity = moles / volume (L)
0.200 = 2.50 x 10^-3 / volume (L)
volume = 0.0125 L
volume = 12.5 mL
volume of H2SO4 needed = 12.5 mL
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