10.94 mL of KOH solution is required to titrate 0.5361 g of potassium hydrogen p

Question

10.94 mL of KOH solution is required to titrate 0.5361 g of potassium hydrogen phthalate (KHP; FW=204.224) to the Phenolphthalein endpoint. If 27.96 mL of this solution is required to titrate 96.34 mL of HCl to the Phenolphthalein endpoint, what is the concentration of the HCl solution?

__________ M

One antacid tablet is ground and dissolved in 50 mL of DI water. Indicator and 100.00 mL of the HCl solution are added. If it requires 23.22 mL of the KOH solution to back titrate the excess HCl to the endpoint, how many moles of acid were neutralized by the antacid tablet?

__________ moles

The neutralizing power of an antacid can be defined as the volume (mL) of 0.1000 N HCl that can be neutralized. What is the neutralizing power of this tablet?

__________ mL

Explanation / Answer

Q1

Q1

mol of KHP = mass/MW = 0.5361 /204.224 = 0.002625 mol of KHP

mol of NaOH = mol o kHP = 0.002625

[KOH] = mol/V = 0.002625/(10.94*10^-3) = 0.2399 = 0.24 M

now..

mmol of base = mmol of acid

mmol of base = MV = 0.24 *27.96 = 6.7104 mmol of base

mmol of Hcl = 6.7104

[HCl] = mmol/mL= 6.7104 /96.35 = 0.06964 M

Q2

mmol of Hcl added = MV = 0.06964*100 = 6.964 mmol

mmol fo KOH backtitrated = MV = 23.22*0.24 = 5.5728

mmol fo HCl reacted = 6.964 -5.5728 = 1.3912 mmol of HCl used

mol = 1.3912*10^-3 = 0.0013912 mol of HCl

Q3

V HCl = mmol/M = 1.3912 / 0.06964 = 19.977 mL --> approx 20 mL

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