Question

if you could write out how you solved so i can understand how to actually do it/where i went wrong thank you so much!

1. A standard solution of NaOH is formed by weighing 4.000 g of NaOH and dissolving it in water to make a total of 200.0 ml of solution. This solution is then used to titrate a solution of HCl(aq) of unknown concentration to an endpoint. The endpoint is reached after the addition of 52.67 ml of the NaOH solution. If the original HCI solution had a volume of 26.45 ml, what was the concentration of the original HCl solution? 2. Given that: NO(g) + 2NO(g) 2NO(g) + 0.(g) + 2NO(g) AH = +57.93 kJ AH = -113.14 kJ Calculate AH® for the following reaction: 2NO(g) + O2(g) → NO.(g)

Answer 1

Question 1

Mass of NaOH = 4 g

Volume of the solution = 200 ml  

Concentration of NaOH = 4 g x 1000 ml / (200ml x 40 g/mol ) = 0.5M

Volume of NaOH used in the titration = 52.67 ml

Moles of NaOH used = 52.67 ml x 0.5 M / 1000 = 0.026335 M

Moles of HCl reacted =  0.026335 M

Volume of HCl = 26.45 ml

Concentration of HCl = 0.026335 x 1000 / 26.45 = 0.9956 M

Hence Concentration of HCl is 0.9956 M.

Question 2

We can modify the equation as shown below

Reverse the first equation

2NO2 (g) ===> N2O4  (g)   ΔH = -57.93 kJ

2NO (g)+ O2 (g)===> 2NO2 (g)  ΔH = -113.14 kJ

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2NO (g)+ O2 (g) ===> N2O4  (g) ΔH = - 171.07 kJ

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ΔH of the above reaction is - 171.07 kJ