Im posting seperate questions so just answer this one

If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so

Leviafan [203]

Answer:

Final pH: 9.49.

Round to two decimal places as in the question: 9.5.

Explanation:

The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.

What's the pKb of base B?

Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.

$\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}$ .

$\text{pOH} = \text{pK}_w - \text{pH}$ .

$\text{pK}_w = 14$ .

$\text{pOH} = 14 - 9.5 = 4.5$

$\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613$ .

What's the new salt-to-base ratio?

The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.

Initial:

$n(\text{B}) = c\cdot V = 1.05 \times 0.5 = 0.525\;\text{mol}$ ;
$n(\text{BH}^{+}) = c\cdot V = 0.750 \times 0.5 = 0.375\;\text{mol}$ .

After adding the HCl:

$n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol}$ ;
$n(\text{BH}^{+}) = 0.375+ 0.005 = 0.380\;\text{mol}$ .

Assume that the volume is still 0.5 L:

$\displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}$ .
$\displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}$ .

What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

$\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991$

$\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49$ .

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

3 0

3 years ago

The balanced chemical equation for the combustion of methane is: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Which of the following sta

Jobisdone [24]

Answer:

Option 2. One mole of methane gas reacts with two moles of dioxygen gas, producing one mole of carbon dioxide gas and two moles of gaseous water.

Explanation:

The clue is in the stoichiometry of the reaction:

1 CH₄(g) + 2 O₂(g) → 1 CO₂(g) + 2 H₂O(g)

In any chemistry reaction, the stoichiometric coefficients are moles; moles of reactants to produce moles of products

5 0

4 years ago

11. Write the balanced equation for the reaction of HC2H3O2 with Al(OH)3 to form H2O and AI(C2H3O2)3

TiliK225 [7]

Answer:

3HC2H3O2(aq) + Al(OH)3(aq) --> AI(C2H3O2)3(aq) + 3H2O(l)

Explanation:

HC2H3O2 is the chemical formula for Ethanoic acid which can be written as CH3COOH.

Hence, the balanced equation is stated as 3CH3COOH(aq) + Al(OH)3(aq) --> AI(CH3COO)3(aq) + 3H2O(l)

Acid + base → Salt + Water.

The equation is a neutralization reaction in which the acid, aqeous CH3COOH reacts completely with an appropriate amount of base, aqueous Al(OH)3 to produce salt, aqueous AI(CH3COO)3(aq)

and water, liquid H2O only.

During this reaction, the hydrogen ion, H+, from the Ethanoic acid is neutralized by the hydroxide ion, OH-, from the Aluminum hydroxide to form the water molecule, H2O and aluminium ethanoate.

Thus, it is called a neutralization reaction.

4 0

3 years ago

Which statement goes against the kinetic theory of gases?

AURORKA [14]

the answer is c. Gas molecules will never collide with the walls of the container

8 0

4 years ago

Read 2 more answers

How many grams of NaCl should be obtained to make 150 mL of 4.5 M solution

nadya68 [22]

Answer:

39,5 grams should be obtained of NaCl

Explanation:

We calculate the weight of 1 mol of NaCl:

Weight 1 mol NaCl= Weight Na + Weight Cl= 23g + 35,5g=58,5 g/mol

4,5M--> 4,5 moles NaCl in 1000ml (1L) of solution

1000ml-----4,5 moles NaCl

150 ml------x=(150 mlx4,5 moles NaCl)/1000ml=0,675 moles NaCl

1 mol NaCl--------------58,5 grams

0,675molesNaCl---x= (0,675molesNaClx58,5 grams)/1 mol NaCl

x= 39,4879 grams

3 0

3 years ago