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lina2011 [118]
3 years ago
14

EZ MODE -_- What could you do in an experiment to get more accurate results?  Change two of your variables. Only make observatio

ns that involve sight. Only use new materials. Perform repeated trials​
Chemistry
1 answer:
DENIUS [597]3 years ago
3 0

Answer:

they change the object hehehehe

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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
Please help me it is due in 10 min <br> there is 8 questions
m_a_m_a [10]

Answer:

17. Pluto (probably)

11. C

12. B

13. D

14. A

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3 years ago
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