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Suppose a student adds 25.00 mL of 1.015 M HCl to a 1.50 g antacid tablet. The student boils and then titrates the resulting solution to the endpoint with 0.5015 M NaOH. The titration requires 21.7 mL NaOH to reach the endpoint. How many moles of HCl were neutralized by the NaOH

Answer :

Answer:

Explanation:

Volume 0.5015 M NaOH required to neutralise HCl = 21.7 ml

21.7 ml of .5015 M NaOH = 21.7 x .5015 ml of  M NaOH

=10.8826 ml of M NaOH

10.8826 ml of M NaOH will contain  10.8826 / 1000 moles of NaOH

= 10.88 x 10⁻³ moles  of NaOH

So moles of HCl neutralised = 10.88 x 10⁻³ moles Ans

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