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igor_vitrenko [27]
3 years ago
13

Perform the following

Chemistry
1 answer:
goblinko [34]3 years ago
5 0

Answer:

8.9 or 8.979

Explanation:

When you add or subtract, you assign significant figures in the answer based on the number of decimal places in each original measurement.

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Calculate the pH of a buffer solution created by reacting 100 mL of 0.1 M NH3 with 90 mL of 0.1 M HNO3. (Remember, you can find
algol13

This question is asking for the pH of a buffer solution between ammonia and nitric acid, with given volumes and concentrations. At the end, the result turns out to be 10.488.

<h3>Buffers</h3>

In chemistry, buffers are known as substances attempting to hold a relatively constant pH by mixing and acid and a base (weak and strong). In such a way, for the substances given, the first step will be to calculate the consumed moles as they are mixed:

n_{NH_3}=0.1L*0.1mol/L=0.01mol\\\\n_{HNO_3}=0.09L*0.1mol/L=0.009mol

Now, since ammonia is in a greater proportion, one can calculate how much of it is left after being consumed by the nitric acid:

n_{NH_3}^{left}=0.01mol-0.009mol=0.001mol

And its new concentration:

[NH_3]=\frac{0.001mol}{0.1L+0.09L} =0.00526M

Next, with ammonia's ionization:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

We set up the equilibrium expression based on ammonia's Kb:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}

Which can be solved by introducing x and using ammonia's Kb:

1.8x10^{-5}=\frac{x^2}{0.00526M}\\ \\

Then, we solve for x which is also equal to the concentration of ammonium and hydroxide ions in the solution:

x=\sqrt{0.00526*1.8x10^{-5}}=0.000308M

Ultimately, we calculate the pOH and then turn it into pH with:

pOH=-log(0.00308)=3.512\\\\pH=14-3.512=10.488

Learn more about buffers: brainly.com/question/24188850

6 0
2 years ago
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