Question

Calculate the heat when 150 ml of .500 M HCI is mixed with450 ml of .100 M barium hydroxide. Assuming the temperature of both solutions was initially 25℃ and that the final mixture has a mass of 600 g and specific heat if water, calculate the final temperature of the mixture?

Answer 1

Answer on Question #43895 - Chemistry - Other

Question:

Calculate the heat when $150\mathrm{ml}$ of .500 M HCl is mixed with $450\mathrm{ml}$ of .100 M barium hydroxide. Assuming the temperature of both solutions was initially $25^{\circ}\mathrm{C}$ and that the final mixture has a mass of $600\mathrm{g}$ and specific heat if water, calculate the final temperature of the mixture?

$$c _ {i} (\mathrm {B a} (\mathrm {O H}) _ {2}) = 0. 1 0 0 \frac {m o l}{L}$$ $$V (\mathrm {B a} (\mathrm {O H}) _ {2}) = 4 5 0. 0 m L$$ $$c _ {i} (\mathrm {H C l}) = 0. 5 0 0 \frac {m o l}{L}$$ $$V (\mathrm {H C l}) = 1 5 0. 0 m L$$ $$m (\text {s o l u t i o n}) = 6 0 0. 0 g$$ $$t _ {1} (\mathrm {H C l}) = t _ {1} (\mathrm {B a} (\mathrm {O H}) _ {2}) = 2 5 ^ {\circ} \mathrm {C}$$ $$c _ {\mathrm {P}} \left(\mathrm {H} _ {2} \mathrm {O}\right) = 4. 1 8 1 3 \frac {J}{g \cdot K}$$ $$t _ {2} (\text {s o l u t i o n}) -?$$

Solution:

1. Write the reaction as a molecular equation:

$$\mathrm {B a} (\mathrm {O H}) _ {2} (\mathrm {a q}) + 2 \mathrm {H C l} _ {(\mathrm {a q})} \rightarrow \mathrm {B a C l} _ {2} (\mathrm {a q}) + 2 \mathrm {H} _ {2} \mathrm {O} _ {(\mathrm {l})}$$

2. Then the overall ionic equation for the reaction:

$$\mathrm {B a} _ {\mathrm {(a q)}} ^ {2 +} + 2 \mathrm {O H} _ {\mathrm {(a q)}} ^ {-} + 2 \mathrm {H} _ {\mathrm {(a q)}} ^ {+} + 2 \mathrm {C l} _ {\mathrm {(a q)}} ^ {-} \rightarrow \mathrm {B a} _ {\mathrm {(a q)}} ^ {2 +} + 2 \mathrm {C l} _ {\mathrm {(a q)}} ^ {-} + 2 \mathrm {H} _ {2} \mathrm {O} _ {\mathrm {(l)}}$$

3. The net ionic equation for the reaction:

$$2 \mathrm {H} _ {\mathrm {(a q)}} ^ {+} + 2 \mathrm {O H} _ {\mathrm {(a q)}} ^ {-} \rightarrow 2 \mathrm {H} _ {2} \mathrm {O} _ {\mathrm {(l)}}$$

4. When we mix strong acid with strong base, we obtain the heat from the reaction of water synthesis. The standard enthalpy of neutralization for such reaction is constant $\Delta H_{\mathrm{N}}^{\ominus} = -56000\frac{J}{mol}$ . This value shows us the change of enthalpy during water synthesis. We can calculate the heat released from such reaction by the equation:

$$Q = - \frac {\Delta H _ {N} ^ {\ominus}}{n _ {i} (\mathrm {H} _ {2} \mathrm {O})}$$

5. The amount of water can be found from the equation:

$$n _ {i} (\mathrm {H} _ {2} \mathrm {O}) = n _ {i} (\mathrm {H C l}) = \frac {n _ {i} (\mathrm {B a} (\mathrm {O H}) _ {2})}{2}$$

6. We compare the amount of acid and base and find that the base is in excess:

$$n _ {i} (\mathrm {B a (O H)} _ {2}) = c _ {i} (\mathrm {B a (O H)} _ {2}) \cdot V (\mathrm {B a (O H)} _ {2}) = 0.100 \frac {m o l}{L} \cdot \frac {450.0 \, mL}{1000} = 0.045 \, m o l$$ $$n _ {i} (\mathrm {H C l}) = c _ {i} (\mathrm {H C l}) \cdot V (\mathrm {H C l}) = 0.500 \frac {m o l}{L} \cdot \frac {150.0 \, mL}{1000} = 0.075 \, m o l$$

The stoichiometric ratio $\frac{n_i(\mathrm{Ba(OH)}_2)}{n_i(\mathrm{HCl})}$ is $1:2$, so for full acid neutralization we need to have

$$n _ {i} (\mathrm {B a (O H)} _ {2}) = \frac {n _ {i} (\mathrm {H C l})}{2} = \frac {0.075 \, m o l}{2} = 0.0375 \, m o l$$

The stoichiometric ratio $\frac{n_i(\mathrm{HCl})}{n_i(\mathrm{H}_2\mathrm{O})}$ is $1:1$, so

$$n _ {i} (\mathrm {H} _ {2} \mathrm {O}) = n _ {i} (\mathrm {H C l}) = 0.075 \, m o l$$

7. Find heat released after reaction:

$$Q = - \Delta H _ {N} ^ {\ominus} \cdot n _ {i} (\mathrm {H} _ {2} \mathrm {O}) = - (- 56000) \frac {J}{m o l} \cdot 0.075 \, m o l = 4200.0 \, J$$

8. Then we can calculate the temperature difference after the reaction:

$$Q = c _ {\mathrm {P}} (\mathrm {H} _ {2} \mathrm {O}) \cdot m (\mathrm {H} _ {2} \mathrm {O}) \cdot \Delta T$$ $$\Delta T = \frac {Q}{c _ {\mathrm {P}} (\mathrm {H} _ {2} \mathrm {O}) \cdot m (\mathrm {H} _ {2} \mathrm {O})} = \frac {4200.0 \, J}{600 \, g \cdot 4.1813 \, \frac {J}{g \cdot K}} = 1.67 \, K$$ $$\Delta T = \Delta t = 1.67^{\circ} \mathrm{C}$$

9. Find the final temperature of the mixture:

$$\Delta t = t _ {2} (\text {s o l u t i o n}) - t _ {1} (\mathrm {H C l})$$ $$t _ {2} (\text {s o l u t i o n}) = \Delta t + t _ {1} (\mathrm {H C l}) = 25^{\circ} \mathrm{C} + 1.67^{\circ} \mathrm{C} = 26.67^{\circ} \mathrm{C}$$

Answer: The final temperature of the mixture is $26.67^{\circ}\mathrm{C}$.

http://www.AssignmentExpert.com/