What is the concentration of Na+ ions if 25.0 mL of 1.50 M NaOH is reacted with 25.0 mL of 1.50 M
HCl?
I did
M*25ML=1.50M*25.0ML
M=1.50M*25.0ML/25ML
M=1.50 M Na
Is it correct?
No, you have diluted 25.0 mL of 1.50M NaOH to 50.0 mL total volume so (Na^+) is now just 1/2 what it was.
1.5 M Na^+ x 25/50 = ?
To determine the concentration of Na+ ions in the reaction between 25.0 mL of 1.50 M NaOH and 25.0 mL of 1.50 M HCl, you need to first understand the stoichiometry of the reaction. The balanced equation for the reaction is:
NaOH + HCl -> NaCl + H2O
From the equation, you can see that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl. Therefore, the concentration of Na+ ions is equal to the concentration of NaCl produced in the reaction.
To find the concentration of NaCl, you need to use the concept of the mole-to-mole ratio. In this case, the ratio is 1:1 between NaOH and NaCl. This means that the number of moles of NaCl produced is the same as the number of moles of NaOH reacted.
We start by calculating the number of moles of NaOH reacted:
Moles of NaOH = concentration (M) × volume (L) = 1.50 M × 0.025 L = 0.0375 mol NaOH
Since the mole ratio is 1:1, the number of moles of NaCl produced is also 0.0375 mol.
Now, we need to find the concentration of Na+ ions. The concentration is defined as the number of moles divided by the volume in liters.
Concentration of Na+ ions = Moles of NaCl / Total volume (in liters)
The total volume is the sum of the volumes of NaOH and HCl used in the reaction, which is 25.0 mL + 25.0 mL = 0.05 L.
Concentration of Na+ ions = 0.0375 mol / 0.05 L = 0.75 M Na+
Therefore, the concentration of Na+ ions in the reaction is 0.75 M.