Na Sa Bant HCL -> 50g Hao pt Soy

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

Identify the type of bonding molecular geometry and intermolecular forced experienced by the compounds hf and hbr which substanc

Serhud [2]

Both HF and HBr experience H-bond, dipole, and dispersion forces. These forces allow the compounds to have stronger bonds with each other. HF would experience a higher amount of attraction between its molecules. This is because fluorine is more electronegative than bromine.

Hope this helps!

6 0

3 years ago

Please Help I WILL REWARD BRAINLIEST

anastassius [24]

Most earthquakes occur along the edge of the oceanic and continental plates. The earth's crust (the outer layer of the planet) is made up of several pieces, called plates. The plates under the oceans are called oceanic plates and the rest are continental plates. The plates are moved around by the motion of a deeper part of the earth (the mantle) that lies underneath the crust. These plates are always bumping into each other, pulling away from each other, or past each other. The plates usually move at about the same speed that your fingernails grow. Earthquakes usually occur where two plates are running into each other or sliding past each other.

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3 years ago

Describe the displacement method for measuring the volume of an object

jeyben [28]

<span>Volume is a measure of the amount of space an object takes up. When a cylinder is submerged in the water it pushes water out of the way. If you measure the amount the water level increases, you can find the volume of the water pushed out of the way. </span>

6 0

3 years ago

If a sample of sodium chloride with a mass of

alex41 [277]

Original molarity was 1.7 moles of NaCl

Final molarity was 0.36 moles of NaCl

Given Information:

Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity

Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity

1. Solve for the molarity of the original (concentrated) solution.

Molarity (M) = moles of solute (mol) / liters of solution (L)

Convert the given information to the appropriate units before plugging in and solving for molarity.

Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)

2. Solve for the molarity of the final (diluted) solution.

Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.

Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution

Molarity (M) of the final solution = 0.36 M NaCl

I hope this helped:))

6 0

2 years ago

Na Sa Bant HCL -> 50g Hao pt Soy​

Na Sa Bant HCL -> 50g Hao pt Soy