2. A standard solution of 0.342 M NaOH was used to determine the concentration o

Question

2. A standard solution of 0.342 M NaOH was used to determine the concentration of a hydrochloric acid solution. If 46.33 mL of NaOH is needed to neutralize 25.00 mL of the acid, what is the molar concentration of the acid? Show work

3. Assume the rxn is -53.5 kJ/mole, the specific heat of any dilute water solution is 4.18 J/g t, and the density of any dilute water solution is 1.0 g/mL. In a “coffee cup calorimeter” (open system), 3.00 g of sodium hydroxide was added to 50.0 mL of water. Then it was mixed with 75.0 mL of 1.00 M HCl. The initial temperature of both solutions was 21.50C. The final temperature of the mixture was 28.80C. Calculate the joules of heat released in this reaction and the percent error.

Explanation / Answer

2) The reaction is HCl + NaOH ---> NaCl + H2O

Moles of NaOH = M x V ( in liters) = 0.342 x 46.33/1000 = 0.01584486

as per reaction HCl and NaOH react in 1:1 , hence at neutralisation moles of NaoH = moles of Hcl = 0.01584486

vol of HCl used = 25 ml = 0.025 L

we have formula Molarity = moles / volume in L

hence Molarity of HCl = 0.01584486 / 0.025 = 0.6338 M

3) moles of NaoH = mass of NaoH / molar mass of NaOH= 3 / 40 = 0.075

moles of HCl = M x V = ( 1 x 75/1000) = 0.075

hence both NaoH and HCl are in equal ratios

Hence moles of neutralisation = 0.075

heat of reaction = delat H x moles = - ( 53.5) x 0.075 = -4.0125 KJ ( -ve sign indicates heat released)

this is theoretically calculated or expected heat

Now we calculate heat released based on tempearure rise

total solution volume = 50+75 = 125 ml

mass of solution = density x vol = 1 x 125 = 125 g ( dneisty = 1g/ml)

now Heat absorbed by solution = specific heat of solution x temp change x mass

       = ( 4.18 x ( 28.8-21.5) x 125)

   = 3814.25 J = 3.81425 KJ is heat released in reaction.

% error = 100 x ( difference of heat released of expeirmental and theoretical) / ( average value of heats)

   difference in heats = ( 4.0125 - 3.81425) = 0.19825

average of heats = ( 4.0125 + 3.81425 ) /2 = 3.913375 KJ

now % error = ( 100 x 0.19825) / ( 3.913375)

= 5 %

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