mol HCl = 0.150 x 0.0450 = 0.00675
mol NaOH = 0.200 x 0.0300 = 0.00600
...........HCl + NaOH ==>NaCl + H2O
initial..0.00675.0.00600.....0.....0
change..-0.006...-0.006...0.006..0.006
equil....0.00075..........0.006..0.006
You can see that thae HCl is in excess so (HCl) = (H^+) = 0.00075/total volume in liters. Then convert to pH = -log(H^+).
A student mixes 45.0 ml of .150 M HCL with 30.0 ml of .200 M NaOH. What is the pH of the resulting solution?
1 answer
1 answer