Question

Calculate the heat of neutralization when 50 mL of 1 M HCl(aq) is mixed with 50 mL of 1 M NaOH(aq) at a constant temperature of 25 °C. The specific heat capacity of the solution is 4.18 J/g·°C, and the density of the solution is 1.00 g/mL.

Answer 1

To calculate the heat of neutralization, we need to consider the balanced chemical equation for the reaction between HCl and NaOH:HCl aq  + NaOH aq   NaCl aq  + H2O l From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water. First, let's determine the number of moles of HCl and NaOH in the solution. Since both solutions have the same concentration and volume, they will react completely with each other.moles of HCl =  concentration of HCl    volume of HCl moles of HCl =  1 mol/L    50 mL    1 L/1000 mL moles of HCl = 0.05 molmoles of NaOH =  concentration of NaOH    volume of NaOH moles of NaOH =  1 mol/L    50 mL    1 L/1000 mL moles of NaOH = 0.05 molNow, we need to find the heat of neutralization for the reaction. The heat of neutralization is the heat released when 1 mole of water is formed from the reaction between an acid and a base. The heat of neutralization for strong acids and strong bases, like HCl and NaOH, is approximately -57.32 kJ/mol.Since 0.05 moles of water are formed in this reaction, we can calculate the heat released:q =  moles of water    heat of neutralization q =  0.05 mol    -57.32 kJ/mol q = -2.866 kJThe heat of neutralization when 50 mL of 1 M HCl aq  is mixed with 50 mL of 1 M NaOH aq  at a constant temperature of 25 C is -2.866 kJ.