posted by Anna .
How many mL of 18 M H2SO4 are required to react with 250 mL of 2.50 M Al(OH)3 if the products are aluminum sulfate and water? How do i solve this?
1.Write the equation and balance it.
2.Convert 250 mL of 2.50 Al(OH)3 to moles. moles = M x L.
3. Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles H2SO4.
4. Convert moles H2SO4 to volume. moles = M x L. Solve for L and convert to mL.
i. calculate the pH of the .833 sodium acetate solution
ii. Calculate the pH after adding .700 grams of NaOH to the 35 ml of .833 M sodium acetate solution
iii. Calculate the pH after adding 1.5 mL of 12 M HCl to the original 35 mL of .833 M sodium acetate solution
250mL x 1L/1000mL x 2.50mol Al(OH)3/1L x 3 mol H2SO4/2 mol Al(OH)3 x 1L/18.0 mol H2SO4 x 1000mL/1L = 52.1mL H2SO4