0.8 Molar NaOH is added to 30 mL of HCI and the volume of base added vs. pH is graphed which is 7. What molarity is the HCI solution? This is to tell how many moles of sodium hydroxide you were adding.
To determine the molarity of the HCl solution, we can use the concept of stoichiometry.
First, let's establish the balanced chemical equation for the reaction between NaOH (sodium hydroxide) and HCl (hydrochloric acid):
NaOH + HCl -> NaCl + H2O
From the equation, we can see that the ratio of NaOH to HCl is 1:1. This means that for every mole of NaOH used, one mole of HCl is reacted.
In the given problem, the volume of NaOH (base) added vs. pH is graphed until the pH reaches 7. This indicates that the point of neutralization has been reached, and all the HCl has reacted with NaOH.
Since 0.8 M NaOH was added, we can calculate the number of moles of NaOH used:
Number of moles NaOH = Molarity × Volume (in liters)
= 0.8 mol/L × 0.030 L
= 0.024 moles
Since the stoichiometric ratio between NaOH and HCl is 1:1, the number of moles of HCl present in the reaction is also 0.024 moles.
To determine the molarity of the HCl solution, divide the number of moles of HCl by the volume of the HCl solution:
Molarity = Number of moles HCl / Volume (in liters)
= 0.024 moles / 0.030 L
= 0.8 M
Therefore, the molarity of the HCl solution is 0.8 M.