This is my last question, as I have been asking a few. You wish to raise a solution from room temperature (20 degreeC) to boiling (100 degreeC). If you mix equal volumes of HCl and NaOH, both of equal concentration, what must the concentration be? (Assume specific heat of solution is close to that of water, 4.18 J/gdegreeC, and the heat of neutralization for strong acids with strong bases is 55.9 kJ/mol.
I am just completely lost without it giving me volume. I don't know how to go about calculating molarity without anything telling me about moles or liters, etc. Thanks for any help, I appreciate it.
I am just completely lost without it giving me volume. I don't know how to go about calculating molarity without anything telling me about moles or liters, etc. Thanks for any help, I appreciate it.
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You don't really need a volume, as it's all about concentrations; you're able to figure it out with an arbitrary volume. You can take, for example, 1 g of water; for tthat to reach the desired temperature, you need
1 g * 4.18 J/(g °C) * 80 °C = 334.4 J
Now, you just need to figure aout how many moles of strong acid or base need to react to produce that amount of heat:
334.4 J / 55.9 kJ/mol = 5.98 * 10^-3 mol
This is the amount of strong base or acid you need to bring 1 g of water to boil if you add the fitting amount of base or acid WITHOUT water - e.g. by adding neat sodium hydroxyde to aqueous HCl, [neglecting enthalpy of solvation]. So, to reach that temperature when combining two solutions, we obviously need to double the concentrations/amounts of substance in the sample; thus, the concentration of acid or base needs to be:
2 * 5.98 * 10^-3 mol / 0.001 L = 11.96 M
when we assume a density of 1 g/mL.
Now, there is a problem with that calculation, I'm afraid: At such high concentrations (a solution of HCl, for example, is concentrated at about 12 M), the properties of the acid or base solutions clearly differ from those of pure water - and that includes boiling point, density, heat capacity etc., so the calculation above is only a very crude estimate....
1 g * 4.18 J/(g °C) * 80 °C = 334.4 J
Now, you just need to figure aout how many moles of strong acid or base need to react to produce that amount of heat:
334.4 J / 55.9 kJ/mol = 5.98 * 10^-3 mol
This is the amount of strong base or acid you need to bring 1 g of water to boil if you add the fitting amount of base or acid WITHOUT water - e.g. by adding neat sodium hydroxyde to aqueous HCl, [neglecting enthalpy of solvation]. So, to reach that temperature when combining two solutions, we obviously need to double the concentrations/amounts of substance in the sample; thus, the concentration of acid or base needs to be:
2 * 5.98 * 10^-3 mol / 0.001 L = 11.96 M
when we assume a density of 1 g/mL.
Now, there is a problem with that calculation, I'm afraid: At such high concentrations (a solution of HCl, for example, is concentrated at about 12 M), the properties of the acid or base solutions clearly differ from those of pure water - and that includes boiling point, density, heat capacity etc., so the calculation above is only a very crude estimate....