I couldn\'t not figure out how to do part !!B!!! Of number 7 could someone help

Question

I couldn't not figure out how to do part !!B!!! Of number 7 could someone help me find the concentration of the ions ? (7) In a coffee-cup calorimeter. 100.0 mL of 0.50M HCI is added to 40.0mL of 1.00M NaOH to make 140.0 g of solution at an initial temperature of 28.2°C. 130 pts a) If the enthalpy for the reaction between a strong acid and a strong base is -56 kJ/mol, calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.18 J/g C and assume no heat loss to the surroundings. hnle 9: 140,0g(54-415) (28.2 262 +273 k- 1101

Explanation / Answer

Solution

Part B

Given data

Initial concentrations and volume

100.0 ml of 0.50 M HCl

40.0 ml of 1.00 M NaOH

What are the concentrations of the ions after the reaction.

Reaction equation

HCl (aq)   + NaOH(aq) --------- > H2O(l) + NaCl(aq)

Now lets calculate the moles of the HCl and NaOH using their concentrations and volumes

Moles = molarity * volume in liter

Moles of HCl = 0.50 mol per L * 0.100 L = 0.05 mol HCl

Moles of NaOH = 1.00 mol per L * 0.040 L = 0.04 mol NaOH

Since moles of the NaOH are less than moles of HCl therefore the NaOH is the limiting reactant and therefore used up completely

So the moles of HCl remain after reaction = 0.05 mol – 0.04 mol = 0.01 mol HCl

NaCl is the soluble salt therefore it remain in the solution.

So after the reaction we will get

0.05 mol Cl- and 0.01 mol H+ also we will have 0.04 mol Na+ in the solution

Therefore final concentration of the each ion is calculated as follows at total volume

Total volume = 100.0 ml + 40.0 ml = 140.0 ml = 0.140 L

Molarity = moles / volume in liter

Concentration of the [Cl- ]= (0.05 mol /0.140 L) = 0.357 M

Concentration of [Na+] = (0.04 mol / 0.140 L) = 0.2857 M

Concentration of the [H+] = (0.01 mol / 0.140 L) = 0.07143 M

So these are the concentrations of the each ions after the reaction.

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