If 30.0ml of 0.500 M KOH are needed to neutralize 10.0mL of HC1 of unknown concentration (molarity), what is?
moles KOH = 0.0300 L x 0.500 M= 0.0150 KOH + HCl = KCl + H2O the ratio between KOH and HCl is 1 : 1 moles HCl = 0.0150 molarity HCl = 0.0150 mol / 0.0100 L =1.50 M Moles H2SO4 = 0.0200 L x 0.100 M = 0.00200 H2SO4 + 2 NaOH = Na2SO4 + 2 H2O the ratio between H2SO4 and NaOH is 1 : 2 moles NaOH needed = 0.00200 x 2 = 0.00400 Volume NaOH needed = 0.00400 / 0.100 M = 0.0400 L => 40.0 mL
What is the molarity of a NaOH solution if 15.0 ml is exactly neutralized by 7.5 ml of a 0.02M HC2H3)2 solutio
M1V1 =M2V2 15 X M1 = 7.5 X .02 M1 = .01
What is the pH of a solution prepared by diluting 3.0 mL of 2.5 M HCl to a final volume of 100 mL with H2O?
Molarity of Acid after dilution = 3 × 2.5 / 100= 7.5 × 10 ^-2As one molecule HCl release one H+ ion, Molarity of H+ ions will be 7.5 × 10^-2pH = -log 7.5 x 10^-2= 1.125
What is the molarity of the H+? If 64.3 mL of 0.440 M HBr is mixed with 13.0 mL of 1.15 M HCl.?
moles HBr = 0.0643 L x 0.440 M=0.0283 moles HCl = 0.0130 L x 1.15 M=0.0150 total moles H+ = 0.0433 total volume = 0.0773 L [H+]= 0.0433 / 0.0773 = 0.560 M moles NaOH = 0.0276 L x 0.120 M=0.00331 moles LiOH = 0.0133 L x 1.13 M=0.0150 total volume = 0.0409 L [OH-]= 0.0150/ 0.0409 =0.367 M
What is the pH of a solution formed by mixing 40mL of 0.10 M HCL with 10mL of 0.45M of NaOH?
No.of equivalent of HCl in 40 ml of 0.1 M HCl =40/1000 X 0.1 = 0.004No.of equivalent of of NaOH in 45 ml of NaOH = 10/1000 X 0.45 = 0.0045So excess of equivalent of NaOH in the total of 50 ml. Of solution = 0.0045–0.0040 =0.0005No of equivalent of NaOHin 1000 ml =0.0005 X1000/50 =0.01 NSo pOH = -log(0.01) = 2pH= 14-pOH =14–2= 12.——————————————————————