Question #bcb2d

1 Answer
Arcuric Acid
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

Related questions
Impact of this question
2266 views around the world
You can reuse this answer
Creative Commons License