# Calculate the enthalpy change for the neutralization of HCl by NaOH?

## In a coffee cup calorimeter, 50.0 mL of 1.0 M NaOH and 50.0 mL of 1.0 M HCl are mixed. Both solutions are originally at 23.5 degrees C and after the reaction, the final temperature is 29.9 degrees Celsius. Calculate the enthalpy change for the neutralization of HCl by NaOH I don't understand how to do this problem when you don't have the densities of the substances? Can you find the grams from the molar mass? Would that be OK?

May 25, 2018

You need some more data....$\text{the specific heat capacity of water...}$

#### Explanation:

I do not have these data at hand.

You interrogate the reaction...

$N a O H \left(a q\right) + H C l \left(a q\right) \rightarrow N a C l \left(s\right) + {H}_{2} O \left(l\right) + \Delta$

You assume the heat capacity of the solution is the SAME as the heat capacity of water. And thus you can measure the enthalpy associated with the formation of $50 \times {10}^{-} 3 \cdot m o l$ of salt....

You can also assume that the densities of the solutions are close (enuff) to $1 \cdot g \cdot m {L}^{-} 1$... And so you got $\Delta T$...and you can find the heat exchanged with $\text{the specific heat capacity of water...}$