HNO3 + NaOH ==> NaNO3 + H2O
mols HNO3 = M x L = approx 0.0025
mols NaOH = M x L = approx 0.0035
excess NaOH available is approx 0.001mols
M NaOH = mols/L = 0.001/0.045 = approx 0.02M
You need to redo all of these calculations since I've estimated.
Then poH = -log(OH^-)
Calculate pOH and substitute into the equation below.
pH + pOH = pKw = 14
You know pKw and pOH, solve for pH.
Is the solution acidic or basic?
pH<7 = acidic
pH=7 - neutral
pH>7 = basic
What is the pH of the resulting solution when 25.0mL of 0.100 HNO3(aq)are reacted with 20.0 mL of 0.175M NaOH(aq)? Is the resulting solution acidic or basic in nature?
2 answers
That was super helpful, thanks!