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Answer:
0.1410 M
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
HCl + NaOH —> NaCl + H₂O
From the balanced equation above,
The mole ratio of the acid, HCl (nA) = 1
The mole ratio of the base, NaOH (nB) = 1
Next, the data obtained from the question. This include:
Volume of acid, HCl (Va) = 25 mL
Volume of base, NaOH (Vb) = 34.55 mL
Concentration of base, NaOH (Cb) = 0.1020 M
Concentration of acid, HCl (Ca) =?
CaVa / CbVb = nA/nB
Ca × 25 / 0.1020 × 34.55 = 1/1
Ca × 25 / 3.5241 = 1/1
Cross multiply
Ca × 25 = 3.5241 × 1
Ca × 25 = 3.5241
Divide both side by 25
Ca = 3.5241 / 25
Ca = 0.1410 M
Therefore, the concentration of the acid, HCl is 0.1410 M
Answer : The value of equilibrium constant for this reaction at 262.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 262.0 K is
Answer:
72.6 mL
Explanation:
A quick way to solve this titration problem when you have a monoprotic acid is to use the Dilution equation, M1V1=M2V2.
.589(x)=.821(52.1)
X=72.6 mL
Answer:
a) pH = 13.176
b) pH = 13
c) pH = 12.574
d) pH = 7.0
e) pH = 1.46
f) pH = 1.21
Explanation:
HBr + NaOH ↔ NaBr + H2O
∴ equivalent point:
⇒ mol acid = mol base
⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)
⇒ Va = 0.025 L
a) before addition acid:
⇒ <em>C </em>NaOH = 0.150 M
⇒ [ OH- ] = 0.150 M
⇒ pOH = - Log ( 0.150 )
⇒ pOH = 0.824
⇒ pH = 14 - pOH
⇒ pH = 13.176
b) after addition 5mL HBr:
⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M
⇒ <em>C </em>HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M
⇒ [ OH- ] = 0.1 M
⇒ pOH = 1
⇒ pH = 13
c) after addition 15mL HBr:
⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M
⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M
⇒ [ OH- ] = 0.0375 M
⇒ pOH = 1.426
⇒ pH = 12.574
d) after addition 25mL HBr:
equivalent point:
⇒ [ OH- ] = [ H3O+ ]
⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²
⇒ [ H3O+ ] = 1 E-7
⇒ pH = 7.0
d) after addition 40mL HBr:
⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M
⇒ [ H3O+ ] = 0.035 M
⇒ pH = 1.46
d) after addition 60mL HBr:
⇒ <em>C</em> HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M
⇒ [ H3O+ ] = 0.062 M
⇒ pH = 1.21
