Question

The neutralization of H3PO4 with KOH is exothermic. H3PO4(aq)+3KOH(aq)⟶3H2O(l)+K3PO4(aq)+173.2 kJ If 60.0 mL of 0.200 M H3PO4 is mixed with 60.0 mL of 0.600 M KOH initially at 23.43 °C, predict the final temperature of the solution, assuming its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C). Assume that the total volume is the sum of the individual volumes.

Answer 1

Answer:

Final temperature of solution is 27.48$^{0}\textrm{C}$

Explanation:

Total volume of mixture = (60.0+60.0) mL = 120.0 mL

We know, density = (mass)/(volume)

So mass of mixture = $(120.0\times 1.13)g=135.6 g$

Amount of heat released per mol of $H_{3}PO_{4}$ = $\frac{(m_{mixture}\times C_{mixture}\times \Delta T_{mixture})}{n_{H_{3}PO_{4}}}$

Where, m represents mass , C represents specific heat, $\Delta T$ represents change in temperature and n is number of moles

As this reaction is an exothermic reaction therefore temperature of mixture will be higher than it's initial temperature.

Let's say final temperature of mixture is T $^{0}\textrm{C}$

So, $\Delta T_{mixture}=(T-23.43)^{0}\textrm{C}$

Here $m_{mixture}=135.6 g$ and $C_{mixture}=3.78J/(g.^{0}\textrm{C})$

Moles of H_{3}PO_{4} are added = $\frac{0.200}{1000}\times 60.0moles$ = 0.012 moles

So, $(173.2\times 10^{3})J=\frac{[(135.6g)\times (3.78J.g^{-1}.^{0}\textrm{C}^{-1})\times (T-23.43)^{0}\textrm{C}]}{0.012}$

or, T = 27.48$^{0}\textrm{C}$

So, final temperature of solution is 27.48$^{0}\textrm{C}$