Question

The neutralization of H 3 PO 4 with KOH is exothermic. H 3 PO 4 ( aq ) + 3 KOH ( aq ) ⟶ 3 H 2 O ( l ) + K 3 PO 4 ( aq ) + 173.2 kJ If 55.0 mL of 0.227 M H 3 PO 4 is mixed with 55.0 mL of 0.680 M KOH initially at 22.62 °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

Step-by-step explanation:

It is known that,

No. of moles = Molarity × Volume

So, we will calculate the moles of
`H_(3)PO_(4)` as follows.

No. of moles =
`0.227 * 0.055 L`

= 0.0125 mol

Now, the moles of KOH are as follows.

No. of moles =
`0.680 * 0.055 L`

= 0.0374 mol

And,
`3 * \text{moles of} H_(3)PO_(4)` =
`3 * 0.0125`

= 0.0375 mol

Now, the balanced reaction equation is as follows.

`H_(3)PO_(4)(aq) + 3KOH(aq) \rightarrow 3H_(2)O(l) + K_(3)PO_(4)(aq) + 173.2 kJ`

This means 1 mole of
`H_(3)PO_(4)` produces 173.2 kJ of heat. And, the amount of heat produced by 0.0125 moles of
`H_(3)PO_(4)` is as follows.

M =
`(0.0125 mol * 173.2 kJ)/(1)`

= 2.165 kJ

Total volume of the solution = (55.0 + 55.0) ml

= 110 ml

Density of the solution = 1.13 g/ml

Mass of the solution = Volume × Density

=
`110 ml * 1.13 g/ml`

= 124.3 g

Specific heat = 3.78
`J/g^(o)C`

Now, we will calculate the final temperature as follows.

q =
`mC * \Delta T`

2165 J =
`124.3 * 3.78 * (T - 22.62)^(o)C`

2165 - 469.854 =
`T - 22.62^(o)C`

17.417 =
`T - 22.62^(o)C`

T =
`40.04^(o)C`

Thus, we can conclude that final temperature of the solution is
`40.04^(o)C`.