A 28.39 ml sample of 0.0750 M Ca(OH)2 is added to 31.34ml of 0.100 M HCl. Is the

Question

A 28.39 ml sample of 0.0750 M Ca(OH)2 is added to 31.34ml of 0.100 M HCl. Is the resulting acidic, basic, orneutral? Basic If the solution is not neutral how many moles of excess acidor base are present? .000562 How many additional ml of 0.075M Ca(OH)2 of 0.100 M HCl wouldbe required to neutralize the solution? A 28.39 ml sample of 0.0750 M Ca(OH)2 is added to 31.34ml of 0.100 M HCl. Is the resulting acidic, basic, orneutral? Basic If the solution is not neutral how many moles of excess acidor base are present? .000562 How many additional ml of 0.075M Ca(OH)2 of 0.100 M HCl wouldbe required to neutralize the solution? How many additional ml of 0.075M Ca(OH)2 of 0.100 M HCl wouldbe required to neutralize the solution?

Explanation / Answer

a) Balanced chemical equation :                                                Ca(OH)2 + 2HCl   ------------ >CaCl2 + 2H2O From the equation one mole of the base reacts with twomoles of acid. No.of moles of base = 0.02839 L * 0.0750 M                                 = 0.00212925 moles No.of moles of acid   = 0.03134 L * 0.100 M                                  =0.003134 moles As there are less no.of moles of acid than the required ,solution will be basic . b) Actually one mole of the base reacts with two moles ofacid. For 0.003134 moles of base we need 0.003134 moles / 2moles of acid .                                                         = 0.001567 moles Excess no.of moles = 0.00212925 moles -0.001567 moles                                    =0.000562 moles c) Volume of base       = 0.100M * 0.03134 L * 1 mole / 2 moles * 0.0750 M                                    =0.02089 L                                   = 20.89 ml 20.89 ml is sufficient to neutralise 0.100 M 31.34 mlof acid.

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