Question

# 1) How many moles of acid are there in 17 ml of a 7.1 M HCl...

1) How many moles of acid are there in 17 ml of a 7.1 M HCl solution?

2) 18 ml of a 1.0 M HCl solution was added to a flask and titrated against a solution of 3.8 M NaOH. How many ml of the NaOH solution was required to neutralize the acid?

3) What volume (in liters) of a 2.8 M NaOH solution would neutralize 4.8 moles of H2SO4 (diprotic)?

4) Using the formula M1V1 = M2V2, calculate the new concentration of a solution, if 270 ml of a 1.4 M solution was diluted to 450 ml.

1) moles = Molarity x volume in L

moles = 7.1 x 17/1000

moles = 0.1207

2) first calculate the moles of acid present

moles of HCl = 1.0 x 18/1000

moles of HCl = 0.018

so we need 0.018 moles NaOH to neutralize HCl

0.018 = 3.8 X V

V = 0.00474 L

V = 4.74 mL

4.74 mL NaOH required

3) H2SO4 + 2NaOH -----------> Na2SO4 + 2H2O

4.8 x 2 = 9.6 moles of NaOH required

9.6 = 2.8 x V

V = 3.43 L

V = 3.43 L

3.43 L NaOH required

4) M1V1 = M2V2

M1 = 1.4 M , V1 = 270 mL

M2 = ? V2 = 450 mL

M2 = (M1V1 / V2)

M2 = (1.4 x 270 / 450)

M2 = 0.84 M

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