It takes 37.16 mL of 0.510 M HCL to neutralize 25.00 mL of NaOh. What is the concentration of the NaOH?
there are .03716*0.510 = 0.01895 moles of HCl
Each mole of HCl neutralizes 1 mole of NaOH
So, there were .01895 moles of NaOH in 0.025L of solution. Thus, the solution was
0.01895mol/0.025L = 0.758M
To find the concentration of NaOH, we can use the concept of stoichiometry. In a neutralization reaction between an acid (HCl) and a base (NaOH), the mole ratio between them is 1:1.
First, let's calculate the moles of HCl using the given volume and concentration.
Moles of HCl = Volume of HCl × Concentration of HCl
= 37.16 mL × 0.510 M
= 18.9456 mmol (millimoles)
Since the mole ratio between HCl and NaOH is 1:1, we know that the number of millimoles of NaOH is also 18.9456 mmol.
Now, we can find the concentration of NaOH by dividing the moles of NaOH by the volume of NaOH.
Concentration of NaOH = Moles of NaOH / Volume of NaOH
= 18.9456 mmol / 25.00 mL
= 0.7578 M
Therefore, the concentration of NaOH is 0.7578 M.