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2 years ago

9

The buffer is prepared by adding 250mL of 0.80M NH3 to 150mL of 0.50M NH4NO3. What is the pH of the final solution (Kb for NH3 =

1.8 x 10^-5)

1 answer:

madam [21]2 years ago

6 0

The pH of the solution is obtained to be 9.67.

<h3>What is a buffer?</h3>

A buffer is a solution that serves to militate against changes in acidity or alkalinity. The buffer is often made of a solution of a strong acid and its salt or a weak base and its salt.

Now we have the that;

Number of moles of base = 250 /1000 * 0.80 = 0.2 moles

Number of moles of salt = 150/1000 * 0.50 = 0.075 moles

Total volume of the solution = 250 + 150 = 400 mL or 0.4 L

Molarity of the base = 0.2 moles/0.4 L = 0.5 M

Molarity of the salt = 0.075 moles/ 0.4 L = 0.1875 M

pKb = - log( 1.8 x 10^-5) = 4.74

pOH = pKb + log [salt/base]

pOH = 4.74 + log (0.1875/ 0.5)

pOH = 4.33

pH = 14 - 4.33

pH = 9.67

The pH of the solution is obtained to be 9.67.

Learn more about pH:brainly.com/question/15289741

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Explanation:

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What is the maximum amount of HCl, in grams, that can be produced if 37.5 g of BCl3 and 60.0 g of H2O are reacted according to t

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Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.

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4 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

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Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
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Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago