Since water has a density of about 1g/mL, and we are assuming HCl and NaOH are mostly water, the total mass of the 2 solutions is 100 g.
Due to the conservation of mass, the mass of product must equal the total mass of reactant, therefore the final mass must also be 100 g.
Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
= ![\frac{[SO42-] [H3O+]}{[HSO4-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO42-%5D%20%5BH3O%2B%5D%7D%7B%5BHSO4-%5D%7D)
0.011 = 
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x = 
/ 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x = ![\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }](https://tex.z-dn.net/?f=%5Csqrt%5B-%28-o.111%29%5D%7B%28-0.111%29%5E2%20-%204%28-1%29%20%280.0011%29%20%7D)
/ 2(-1)
x = 0.111 +,- 
/ -2
x = 0.111 +,- 
/ -2
x = 
x = 
, x = 
x = 
, x = 
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963
Answer:
c. 0.2 M HNO₃ and 0.4 M NaF
.
Explanation:
A buffer is defined as the mixture of a weak acid with its conjugate base or a weak base with its conjugate acid.
A weak acid or weak base are defined as an acid or base that partially dissociates in aqueous solution. in contrast, a strong acid or base are acids or bases that is dissociated completely in water.
Thus:
a. 0,2M HNO₃ and 0.4 M NaNO₃. This is a mixture of a strong acid with its conjugate base. <em>IS NOT </em>a buffer.
b. 0.2 M HNO₃ and 0.4 M HF
. This is a mixture of two strong acids. <em>IS NOT </em>a buffer.
c. 0.2 M HNO₃ and 0.4 M NaF
. NaF is the conjugate base of a weak acid as HF is.
The reaction of HNO₃ with NaF is:
HNO₃ + NaF → HF + NaNO₃
That means that in solution you will have a weak acid (HF) with its conjugate base (NaF). Thus, this mixture <em>IS </em>a buffer.
d. 0.2 M HNO₃ and 0.4 M NaOH. This is the mixture of a strong acid with a strong base, thus, this <em>IS NOT </em>a buffer.
I hope it helps!
Fluorine, Chlorine, Bromine, Iodine and Astatine are representing the family of elements called : halogens.
Halogens:
- all have 7 electrons in their outer shell,
- are very reactive elements,
- form - 1 ions.
Answer: C ) They have same chemical properties.
