Acid and Base Pt. 2?

Question

1. Javier need to titrate 2.87g Mg(OH)2. How many milliliters of 0.128 M HCl will he need to neutralize it?

2. Chim performs a titration using a solution of 2.0M HCl and NaOH with an unknown concentration. If 15.0 mL of the acid is needed to neutralize 25.0mL of the base, what is the molarity of her unknown NaOH solution?

Can you please explain the steps!

Answer 1

the formula which is used for neutralization point is
►N1V1 = N2V2 (1)

N2= normality of acid
N1= normality of base
V2 = volume of acid solution
V1 = volume of base solution

on the other hand we have a formula relating normality to molarity:

►N= M * n (2) (n is the no of acidic H for acids or basic OH for bases)

I think now we have what we need:


1.
at first we find 2.87g Mg(OH)2 is how many moles?

► 2.87 /(58.3) = 0.049 mol
dividing by V1 we have: (0.049 / V1) molar Mg(OH)2

for fing the normality of Mg(OH)2 we should multiply the molarity to the no of basic OH of Mg(OH)2 which is 2;

N1 = (0.049 / V1) x2 = (0.098/V1)N

for HCl we have it's molarity and we are going to find its normality; because it has just one acidic H so the molarity and normality are the same for this acid So;

N2= 0.128 N

using formula (1)

►(0.098/V1) * V1= 0.128 * V2
in the left side V1 will cancel out and;

►V2 = 0.098/0.128 = 0.769 lit = 769 mlit


2.
using equattion (1):
2 * 15 = 25 * N1

► N1 =1.2 N
but this is normality ; ased on equation (2) because NaOH has only one basic OH so its normality and molarity are equal so;
M1= 1.2 M

I hope this helps

Answer 2

To neutralize, u need to EQUATE THE NUMBER OF MOLES for both compounds.
M1V1 = M2V2

1)First write down the equation
Mg(OH)2 + 2HCl --> MgCl2 + 2H2O

1 mole Mg(OH)2 reacts with 2 moles HCl for complete neutralization

(M1V1)Mg(OH)2 / (M2V2)HCl = 1/2
Thus, 2(M1V1)Mg(OH)2 = (M2V2)HCl......eq 1

No. of moles of Mg(OH)2= 2.87/ Mr of Mg(OH)2 = x moles
(Please count the molecular mass, Mr. I don't have the periodic table with me)
From eq 1:
2(x moles) = (0.128M)(V)
(V) = (0.128M)/ x moles
And u will get the volume of HCl

2) It will be the same for this question as well
HCl + NaOH --> NaCl + H2O

1 mole HCl reacts with 1 mole NaOH

(M1V1)HCl = (M2V2) NaOH
(2.0M)(0.015dm3) = (M2)(0.025dm3)
M2 = (2.0M)(0.015dm3)/ (0.025dm3)
And u will get the concentration of NaOH

Answer 3

a) 2*2.87/(58*.128) = 0.773 ml

b) 2*15/25 = 1.2 M