Notice that the concentrations and volumes of reactants and products have been a

Question

Notice that the concentrations and volumes of reactants and products have been adjusted in this experiment so that they can cancel properly when applying Hess’ Law:

Reaction A. NaOH(s) NaOH (100 mL, 0.5 M)(aq)

Reaction B. NaOH(s)+ HCl (100 mL, 0.5 M)(aq)NaCl(100 mL, 0.5 M(aq)+ H2O(l)

Reaction C. NaOH(100 mL, 0.5 M)(aq)+ HCl(100 mL, 0.5 M)(aq)NaCl (100 mL, 0.5 M)(aq)+ H2O(l)

a. For Rxn A prove that the final concentration of NaOH is 0.50 M when 2.0 g NaOH solid is dissolved in 100 mL water.

b. For Rxn B, show that when 2.0 g NaOH solid is reacted with 100 mL 0.50 M HCl the product solution is 100 mL 0.50 M NaCl.

c. For Rxn C, first show that the initial concentrations of NaOH and HCl (before reaction) are both 0.50 M, when 50 mL of 1.0 M NaOH is mixed 50 mL 1.0 M HCl. Then show that the concentration f NaCl is 0.50 M after reaction is complete. Why is the final volume of NaCl solution 100 mL?

Explanation / Answer

solution a.

For reaction A NaOH(s) -------->NaOH(aq)

We are adding 2 gm NaoH Into 100 ml Water. Molecular weight of NaOH=40

So number of moles of NaOH =2/40=0.05

And We know that Molarity=number of moles of Solute/Volume of Solution In L.

=0.05/0.1=0.5M

Solution b).

For reaction A NaOH(s) + HCl(aq)(100ml,0.5M) -------->NaCl(aq) + H2O(l)

2gm.

Number of moles=2/40

t=0 0.05mole 0.5×0.1=0.05mole

t=t 0 0 0.05mole

So Molarity of NaCl(aq) =0.05/0.1=0.5M

Solution c).

For reaction A NaOH(aq) + HCl(aq)(100ml,0.5M) -------->NaCl(aq) + H2O(l)

Before reaction moles of NaOH(aq) =0.05 and volume=0.1L

So concentration=0.05/0.1=0.5M

Similarily Before reaction moles of HCl(aq) =0.05 and volume=0.1L

So concentration=0.05/0.1=0.5M

Now 50 ml ,1M NaOH is added to 50ml 1M HCl Total number of mole of HCl=50mmole

So Mole of NaCl(aq.)=50mmole

so Concentration=50/100=0.5M

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