Question

a volume of 100 ml of 1.00 m hcl solution is titrated with 1.00 m naoh solution. you added the following quantities of 1.00 m naoh to the reaction flask. classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.
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200 ml of 1.00 m naoh 10.0 ml of 1.00 m naoh 5.00 ml of 1.00 m naoh 50.0 ml of 1.00 m naoh 150 ml of 1.00 m naoh 100 ml of 1.00 m naoh
before equivalence point at equivalence point after equivalence point

Explanation:

Step1: Calculate moles of HCl

Moles of \(HCl = M\times V=1.00\ M\times0.100\ L = 0.100\ mol\) (using \(n = C\times V\), where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters)

Step2: Determine moles of NaOH for equivalence point

The reaction is \(HCl + NaOH=NaCl + H_2O\), mole ratio \(1:1\). So moles of \(NaOH\) at equivalence point \(= 0.100\ mol\). Volume of \(NaOH\) at equivalence point \(V=\frac{n}{C}=\frac{0.100\ mol}{1.00\ M}=0.100\ L = 100\ mL\)

Step3: Classify each volume

• Before equivalence point: Volume of \(NaOH<100\ mL\). So \(10.0\ mL\) of \(1.00\ M\ NaOH\), \(5.00\ mL\) of \(1.00\ M\ NaOH\), \(50.0\ mL\) of \(1.00\ M\ NaOH\)
• At equivalence point: Volume of \(NaOH = 100\ mL\). So \(100\ mL\) of \(1.00\ M\ NaOH\)
• After equivalence point: Volume of \(NaOH>100\ mL\). So \(200\ mL\) of \(1.00\ M\ NaOH\), \(150\ mL\) of \(1.00\ M\ NaOH\)

Answer:

• Before equivalence point: \(10.0\ mL\) of \(1.00\ M\ NaOH\), \(5.00\ mL\) of \(1.00\ M\ NaOH\), \(50.0\ mL\) of \(1.00\ M\ NaOH\)
• At equivalence point: \(100\ mL\) of \(1.00\ M\ NaOH\)
• After equivalence point: \(200\ mL\) of \(1.00\ M\ NaOH\), \(150\ mL\) of \(1.00\ M\ NaOH\)