If 100.0 mL of 1.020 M HCl and 50.0 mL of 2.040 M NaOH, each at 24.53°C, are mixed in a Styro a) 1.55 M b) 0.77 M c) 0.67 M d) 1.34 M

After mixing HCl and NaOH, which neutralize each other, there are no remaining moles of either in the solution. Therefore the molarity of the solution is 0 M.

Explanation:

To solve this, you first compute for the number of moles of HCl and NaOH. HCl has moles = Molarity x Volume(L) = 1.020 mol/L x 0.100 L = 0.102 mol. NaOH has moles = Molarity x Volume(L) = 2.040 mol/L x 0.050 L = 0.102 mol.

You then find that HCl and NaOH neutralize each other in a 1:1 ratio, so there are no remaining moles of HCl or NaOH.

Since the question asks for molarity after mixing, it's the total moles of solute divided by the total volume of the solution. In this case, there's no remaining solute after the reaction, but the total volume is the sum of volumes of HCl and NaOH (100 mL + 50 mL = 150 mL or 0.150 L). Thus, the molarity of HCl or NaOH in the solution after reaction is 0 mol / 0.150 L = 0 M.

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