Would 0.12g/cm^3 float in water?

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

What is the anion of the compound Cu(CH3COO)2?

olya-2409 [2.1K]

Answer:

The anion of the compound is b CH3COO-

Explanation:

The compound cupric acetate Cu(CH3COO)2 undergo dissociation to form cupric ion Cu2+ and Acetate anion(CH3COO-)

Cu(CH3COO)2⇒ Cu2+ + CH3COO-

From the above equation it can be stated that the anion of Cu(CH3COO)2 is CH3COO-.

4 0

3 years ago

An aqueous solution of sodium hydroxide is standardized by titration with a 0.170 M solution of perchloric acid. If 28.5 mL of b

Ray Of Light [21]

Answer:

0.095M

Explanation:

HClO4 + NaOH = NaClO4 + H2O

Concentration of acid CA= 0.170M

Concentration of base CB= ???

Volume of base VB= 28.5ml

Volume of acid VA= 16.0ml

Number of moles of acid nA= 1

Number of moles of base nB= 1

From

CAVA/CBVB= nA/nB

CB= CAVAnB/VBnA

CB= 0.170×16.0×1/28.5×1

CB= 0.095M

4 0

3 years ago

Given the following unbalanced chemical reaction: As + NaOH Na3AsO3 + H2 What would be the coefficient of the NaOH molecule in t

Lady_Fox [76]

I believe 2 would be the coefficient

6 0

3 years ago

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Cual es la temperatura de congelacion del agua

Sergio039 [100]

<span>La temperatura de congelación del agua es 0 grados </span>centígrados.

7 0

3 years ago