In a given experiment 36.44 mL of NaOH were required for neutralize a solution of HCl 0.35 M.
Calculate the moles of NaOH present to neutralize the HCl in the titration.
mols = M x L = ?
To calculate the moles of NaOH present in the titration, we need to use the equation:
Moles = Volume (in liters) × Concentration
First, let's convert the volume of NaOH used from milliliters to liters by dividing it by 1000:
Volume = 36.44 mL ÷ 1000 = 0.03644 L
Next, we can directly use the given concentration of HCl to calculate the moles of NaOH. In a neutralization reaction, the moles of acid and base will be equal. So:
Moles of NaOH = Moles of HCl
Moles of HCl = Concentration × Volume
Plugging in the values:
Moles of HCl = 0.35 M × 0.03644 L
Now we have the moles of HCl, which is the same as the moles of NaOH:
Moles of NaOH = 0.35 M × 0.03644 L
Calculating this gives us the moles of NaOH present in the titration.