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.

Final temperature of solution is 27.48

Explanation:

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

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

So mass of mixture =

Amount of heat released per mol of =

Where, m represents mass , C represents specific heat, 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

So,

Here and

Moles of H_{3}PO_{4} are added = = 0.012 moles

So,

or, T = 27.48

So, final temperature of solution is 27.48

2b2t

explanation:

2b2t