Answer:
98.68%
Explanation:
Step 1: Write the balanced equation
KBr + AgNO₃ ⇒ AgBr + KNO₃
Step 2: Calculate the moles corresponding to 814.5 mg (0.8145 g) of AgBr
The molar mass of AgBr is 187.77 g/mol.
0.8145 g × 1 mol/187.77 g = 4.338 × 10⁻³ mol
Step 3: Calculate the moles of KBr needed to produce 4.338 × 10⁻³ moles of AgBr
The molar ratio of KBr to AgBr is 1:1. The moles of KBr needed are 1/1 × 4.338 × 10⁻³ mol = 4.338 × 10⁻³ mol.
Step 4: Calculate the pure mass corresponding to 4.338 × 10⁻³ moles of KBr
The molar mass of KBr is 119.00 g/mol.
4.338 × 10⁻³ mol × 119.00 g/mol = 0.5162 g
Step 5: Calculate the purity of KBr
0.5162 g of KBr are in a 0.5231 g-sample. The purity of KBr is:
P = 0.5162 g/0.5231 g × 100% = 98.68%
Molarity of acid=2.5M
pH=5.1.
ka=?
Now
We need to write an eqn to show the dissociation of the acid
HA + H2O === H3O+ + A-
Writing The Equilibrium(Or Acid dissociation constant) of this reaction
Ka =[H3O+] {A-]/ {HA].
The concept behind this is
concentration of Products divided by those of reactants. Water is not written because its a pure liquid and does not affect the Equilibrium constant.
Now If you have any Idea on ICE tables..
You'd know that the concentration of acid will decrease by 2.5-x
Whilst the products...will increase by x each
Note: This is when the ratio of their Moles are in 1:1
ka= x.x/2.5-x
Since the Moles of A- and H3O+ are in 1:1... Their concentrations at equilibrium will be the same
so
Ka= x²/2.5-x
Now what is x??
x is the Hydrozonium ion concentration.
we can get it from the pH formula
pH= -log (H3O+)
Making H3O+ subject by applying Logarithm Rules
H3O+ = 10^-ph
x=10^-5.1
=7.94x10^-6.
Now back to Ka
Ka= x²/2.5-x
Ka= (7.94x10^-6)²/2.5-(7.94x10^-6)
Ka= (7.94x10^-6)²/2.4999
Ka= 2.52x10^-11.
Was a Fun One
Answer:
Explanation:
Given
Required
Determine the percentage error
First, we need to determine the difference in the measurement
The percentage error is calculated as thus:
<em>approximated</em>
A solution in which water is the solvent