How many ml of .20 M NaOh are required to neutralize 50.0 ml of .41 M HBr? M1 = molarity of HBr = 0.41 M : V1 = volume of HBr = 50.0mL M2 = molarity of NaOH = 0.20 : V2 = volume of NaOH By titration we have M1V1=M2V2 V2=(M1V1)/M2= (0.41*50)/0.20 = 102.5mL NaOH What is the pH of a solution to which 35.0 ml of .20 M HCl and 50.0 ml of .42M NaOH have been combined? Moles = molarity * volume Moles of HCl = 0.50M*(35/1000)L = 0.0175 mol HCl Moles of NaOH = 0.42 M * (50/1000)L=0.021 mol NaOH Moles of NaOH left in the solution = 0.021 - 0.0175 = 0.0035 mol NaOH Molarity of NaOH = moles of NaOH left/ total volume of the solution =0.0034/((50+35)/1000)L = 0.04 M pOH = -log[OH-] = -log(0.04)=1.40 pH + pOH = 14 pH = 14-pOH = 14-1.40 = 12.6 Hope it has help you.... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Solution How many ml of .20 M NaOh are required to neutralize 50.0 ml of .41 M HBr? M1 = molarity of HBr = 0.41 M : V1 = volume of HBr = 50.0mL M2 = molarity of NaOH = 0.20 : V2 = volume of NaOH By titration we have M1V1=M2V2 V2=(M1V1)/M2= (0.41*50)/0.20 = 102.5mL NaOH What is the pH of a solution to which 35.0 ml of .20 M HCl and 50.0 ml of .42M NaOH have been combined? Moles = molarity * volume Moles of HCl = 0.50M*(35/1000)L = 0.0175 mol HCl Moles of NaOH = 0.42 M * (50/1000)L=0.021 mol NaOH Moles of NaOH left in the solution = 0.021 - 0.0175 = 0.0035 mol NaOH Molarity of NaOH = moles of NaOH left/ total volume of the solution =0.0034/((50+35)/1000)L = 0.04 M pOH = -log[OH-] = -log(0.04)=1.40 pH + pOH = 14 pH = 14-pOH = 14-1.40 = 12.6 Hope it has help you.... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!.

1. How many ml of .20 M NaOh are required to neutralize 50.0 ml of .41 M HBr?
M1 = molarity of HBr = 0.41 M : V1 = volume of HBr = 50.0mL M2 = molarity of NaOH =
0.20 : V2 = volume of NaOH By titration we have M1V1=M2V2 V2=(M1V1)/M2=
(0.41*50)/0.20 = 102.5mL NaOH What is the pH of a solution to which 35.0 ml of .20 M HCl
and 50.0 ml of .42M NaOH have been combined? Moles = molarity * volume Moles of HCl =
0.50M*(35/1000)L = 0.0175 mol HCl Moles of NaOH = 0.42 M * (50/1000)L=0.021 mol
NaOH Moles of NaOH left in the solution = 0.021 - 0.0175 = 0.0035 mol NaOH Molarity of
NaOH = moles of NaOH left/ total volume of the solution =0.0034/((50+35)/1000)L = 0.04 M
pOH = -log[OH-] = -log(0.04)=1.40 pH + pOH = 14 pH = 14-pOH = 14-1.40 = 12.6 Hope it
has help you.... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Solution
How many ml of .20 M NaOh are required to neutralize 50.0 ml of .41 M HBr?
M1 = molarity of HBr = 0.41 M : V1 = volume of HBr = 50.0mL M2 = molarity of NaOH =
0.20 : V2 = volume of NaOH By titration we have M1V1=M2V2 V2=(M1V1)/M2=
(0.41*50)/0.20 = 102.5mL NaOH What is the pH of a solution to which 35.0 ml of .20 M HCl
and 50.0 ml of .42M NaOH have been combined? Moles = molarity * volume Moles of HCl =
0.50M*(35/1000)L = 0.0175 mol HCl Moles of NaOH = 0.42 M * (50/1000)L=0.021 mol
NaOH Moles of NaOH left in the solution = 0.021 - 0.0175 = 0.0035 mol NaOH Molarity of
NaOH = moles of NaOH left/ total volume of the solution =0.0034/((50+35)/1000)L = 0.04 M
pOH = -log[OH-] = -log(0.04)=1.40 pH + pOH = 14 pH = 14-pOH = 14-1.40 = 12.6 Hope it
has help you.... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!