Determine the volume in milliliters of a 0.162 M NaOH solution needed to neutralize a 235 mL solution of 0.108 M HCl and 0.216 M H2SO4.
157, 470, 235, 940, 156, and 313 mL are all wrong :(( idk why help please!
it was 782 mL. you add 156 + (313 x2) =782, the 156 and 313 are from mv=mv. 313 is multiplied by two bc 2NaOH+H2SO4⟶Na2SO4+2H2O, so it reflects the moles i think
To determine the volume of a solution needed to neutralize another solution, you can use the concept of stoichiometry. In this case, we can assume that the reaction between NaOH (sodium hydroxide) and HCl (hydrochloric acid) is a 1:1 ratio.
Here are the steps to calculate the volume of the NaOH solution:
Step 1: Write down the balanced chemical equation for the reaction.
NaOH + HCl → NaCl + H2O
Step 2: Calculate the number of moles of HCl in the 235 mL solution.
Molarity (M) = moles/volume (L)
moles of HCl = Molarity × volume in liters
moles of HCl = 0.108 M × 0.235 L (convert 235 mL to liters by dividing by 1000) = 0.02538 mol
Step 3: Since the reaction is a 1:1 ratio, the number of moles of NaOH needed will be the same as the moles of HCl.
moles of NaOH = 0.02538 mol
Step 4: Calculate the volume of the 0.162 M NaOH solution.
Moles = Molarity × Volume in liters
Volume in liters = Moles / Molarity
Volume in liters = 0.02538 mol / 0.162 M = 0.1569 L
Step 5: Convert the volume from liters to milliliters.
Volume in milliliters = Volume in liters × 1000
Volume in milliliters = 0.1569 L × 1000 = 156.9 mL
Therefore, the volume in milliliters of a 0.162 M NaOH solution needed to neutralize the 235 mL solution of 0.108 M HCl and 0.216 M H2SO4 is approximately 156.9 mL.