The change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added is 4.75
when the same amount 0.010 mole of NaOH was added to 1.00 L of water the pH = 12
Explanation:
given that:
concentration of acetic acid = 0.50 M
Concentration of base sodium acetate = 0.50 M
ka = 1.8 x 10^-5)
pka = -log [ka]
pka = 4.74
From Henderson-Hasselbalch Equation:
pH = pKa + log ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
pH = 4.74 + Log ![\frac{[0.5]}{[0.5]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.5%5D%7D%7B%5B0.5%5D%7D)
pH = 4.74 + 0
pH = 4.74
Number of moles of NaOH = 0.010 moles
volume 1 litre
molarity = 0.010 M
Moles of acetic acid and sodium acetate before addition of NaOH
FORMULA USED:
molarity = 
acetic acid,
0.5 = number of moles
0.5 is the number of moles of sodium acetate.
number of moles of NaOH 0.010 moles
NaOH reacts in 1:1 molar ratio with acetic acid so
number of moles in acetic acid = 0.5 - 0.010 = 0.49
number of moles in sodium acetate = 0.5 +0.010 = 0.51
new pH
pH = pKa + log
pH= 4.74 + log[0.51] - log[0.49]
pH= 4.75
PH of NaOH of 0.01 M (BASE)
pOH = -Log[0.01]
pOH = 2
pH can be calculated as
14= pH +pOH
pH= 14-2
pH = 12
Answer is: mass fo ammonium chloride is 93.625 grams.
V(NH₄Cl) = 5 L.
c(NH₄Cl) = 0.35 M.
n(NH₄Cl) = V(NH₄Cl) · c(NH₄Cl).
n(NH₄Cl) = 5 L · 0.35 mol/L.
n(NH₄Cl) = 1.75 mol.
M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.
m(NH₄Cl) = n(NH₄Cl) · M(NH₄Cl).
m(NH₄Cl) = 1.75 mol · 53.5 g/mol.
m(NH₄Cl) = 93.625 g.
answer: it is true because a hypothesis is something you think will happen during a experiment
explained: I took the test
Word equation: magnesium chloride + bromine -> magnesium bromide + chlorine
balance equation: MgCl2 + 2Br -> MgBr2 + 2Cl
Complete Question
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?
Answer:
The pK_a value is 
Explanation:
From the question we are told
The volume of base is 
The pH of solution is 
The concentration of the acid is 
From the pH we can see that the titration is between a strong base and a weak acid
Let assume that the the volume of acid is 
Generally the concentration of base
Substituting value
When 13mL of the base is added a buffer is formed
The chemical equation of the reaction is
Now before the reaction the number of mole of base is
![No \ of \ moles[N_B] = C_B * V_B](https://tex.z-dn.net/?f=No%20%5C%20of%20%5C%20moles%5BN_B%5D%20%20%3D%20%20C_B%20%2A%20V_B)
Substituting value
Now before the reaction the number of mole of acid is
Substituting value
Now after the reaction the number of moles of base is zero i.e has been used up
this mathematically represented as
The number of moles of acid is
The pH of this reaction can be mathematically represented as
![pH = pK_a + log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%20%3D%20pK_a%20%2B%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
Substituting values
