A 15.3 mL sample of 0.926 M NaOH is mixed with 23.1 mL of 0.919 M HCl in a coffe

Question

A 15.3 mL sample of 0.926 M NaOH is mixed with 23.1mL of 0.919 M HCl in a coffee-cup calorimeter. The enthalpy of the reaction, written with the lowest whole-number coefficients, is -55.8 kJ. Both solutions are at 18.3°C prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is 1.00 g/mL, the specific heat of all solutions is the same as that of water, and volumes are additive.

Explanation / Answer

V = 15.3 ml

M = 0.926 NaOH

V2 = 23.1 ml

M = 0.919 HCl

Hrxn = -55.8 kJ

T1 = 18.3ªC

Find final temperature

Assume density to be that of water

Cp = 4.184 J/gC

Total Volume:

VT = 15.3+23.1 = 38.4 ml

mass = D*V = 1g/ml * 38.4 ml = 38.4 g of solution

We can assume then, if adibatic:

Q = m*Cp*(Tf-Ti)

We need Q

Q = Hrxn

NaOH + HCl ---> H2O + NaCl Hrnx = -55.8 kJ

Lets see how much we react, lets find limiting reactant

mmoles of NaOH = M1*V1 = 15.3*0.926 = 14.17 mmol of NaOH

mmoels of HCl = M2*V2 = 23.1*0.919 = 21.23 mmol of HCl

if ratio is 1:1, then we have excess of HCl and limitng reactant is NaOH

therefore Hrxn must be based on 14.17 mmol

-55.8 kJ --> 1 mol

then how much for 14.17 mmol or 14.17*10^-3

55.89 kJ*(14.17*10^-3) = 0.7919 kJ or 791.96 J

Now we can go to the Q equation

Q = m*Cp*(Tf-Ti)

791.96 J = 38.4g*4.184J/gC*(Tf-18.3ªC)

solve for Tf

Tf = 791.96J/(38.4g*4.184J/gC) + 18.3

Tf = 4.9292 + 18.3 = 23.229 ºC

Tf = 23.229 ºC

that will be the final temperature given that concentration of reactants

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