Question #bcb2d

1 Answer
Jun 5, 2015

calculate moles of Acid, calculate moles of Base. subtract, OH^- should be in excess, calculate Molarity of OH^- , pOH = -log(OH^-), then 14-pOH = pH

LxM=moles 0.050L x 0.200M NaOH = 0.0100mol

0.030L x 0.200M HNO_3 = 0.006

0.0100mol of OH^- minus 0.006mol of H = 0.004 mole of OH^-in excess
M=mol/L
0.004mol/0.080L
= 0.050M of OH^-

pOH = -log(0.05) = 1.30, 14-1.30= 12.70