How many moles of calcium hydroxide are needed to neutralize 250.0 mL of hydrochloric acid with a pH of 6.000?
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
pH = 6 means (H^+) = 1E-6 M. I don't know if your prof wants you to ignore the small amount of (H^+) from H2O or not. I will ignore it.
mols HCl = M x L = ?
Looking at the coefficients in the balanced equation, that many mols HCl will need 1/2 that amount of Ca(OH)2.
To determine the number of moles of calcium hydroxide (Ca(OH)2) needed to neutralize the hydrochloric acid (HCl), we need to follow a series of steps:
Step 1: Write the balanced chemical equation for the neutralization reaction between calcium hydroxide and hydrochloric acid:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
Step 2: Convert the volume of hydrochloric acid (HCl) from milliliters (mL) to liters (L).
250.0 mL = 250.0 mL * (1 L / 1000 mL) = 0.250 L
Step 3: Use the molarity of hydrochloric acid (pH = 6.000) to calculate the number of moles of hydrochloric acid.
The molarity of hydrochloric acid (HCl) can be calculated using the formula: pH = -log10[H+]
Since the pH is 6.000, the concentration of H+ ions would be 10^(-pH) = 10^(-6.000) = 1.0 x 10^(-6) M.
Molarity (M) = moles (mol) / volume (L)
1.0 x 10^(-6) M = moles / 0.250 L
moles = 1.0 x 10^(-6) M * 0.250 L = 2.5 x 10^(-7) moles of HCl
Step 4: Use the stoichiometry of the balanced equation to determine the moles of calcium hydroxide required.
From the balanced equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl.
Therefore, moles of Ca(OH)2 = (2/1) * moles of HCl = (2/1) * 2.5 x 10^(-7) moles = 5.0 x 10^(-7) moles of Ca(OH)2
So, we need 5.0 x 10^(-7) moles of calcium hydroxide to neutralize 250.0 mL of hydrochloric acid with a pH of 6.000.