how would you answer this problem? part 2. Chemistry 151 Modern Experimental Che

Question

how would you answer this problem?

part 2.

Chemistry 151 Modern Experimental Chemistry Questions: 1. 495 mg of dried KHP is dissolved in 35 mL of distilled water and titrated with potassium hydroxide (KoH). If it took 22.02 mle of KoH to reach the endpoint, determine the concentration of KOH, Show calculations. 000 mL If you forgot to dry the above KHP sample and it was later determined to contain 5.00% water by weight, what would be the actual concentration of the KoH. taking into account the water content? Show calculations. 2. If you titrate 12.50 mL HCI with 23.37 mL of a 0103 MNaoHao reach the endpoint, determine

Explanation / Answer

Molar Mass of KHP = 204.22 g/mol

No. of moles of KHP = Given mass / Molar Mass

= 0.495 (g) / 204.22C (g / mol)

= 0.00242 mol

Concentration of KHP = Concentration of KOH at the equivalence point ---------- (1)

0.0024 * 35 = C * 22.02

No. of moles of KOH, required to neutralise KHP are = 0.00381 mols

Concentration = 0.00381 mols in (22.02 + 35 = 57.02 mL) * 1000 ml = 0.141 M

If KHP conatins 5 % water by weight, then the amount of water possessed by it is = 5 *495/100

= 24.75 g water

Since density of water = 1 g/L

Therefore, volume of water = 24.75 L = 0.024 mL

Total volume of water = 35 + 0.024 ml = 35.024 mL

According to Equation (1) :

0.0024 * 35.024 = C * 22.02

Moles of KOH required to neutralise KHP are = 0.00382 mols

Concentration = 0.00382 mols in (22.02 + 35.024 = 57.044 mL) * 1000 mL = 0.6696 M

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