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avanturin [10]
3 years ago
5

at 298 k the H+ ion concentration of a aqueous solution is 1.00 x 10-5 M what is the pH of the solution

Chemistry
1 answer:
Basile [38]3 years ago
7 0

Answer:

pH = 5.00

Explanation:

The pH is an unit of concentration widely used in chemistry and quality control of aqueous solutions. It serves to find the security of a product to human or pet consumption. Is defined as:

pH = -log [H+]

That means, the pH of a solution that is [H+] = 1.00x10⁻⁵M:

<h3>pH = 5.00</h3>
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Which condition must be met in order for an equation to be balanced?
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Read 2 more answers
Why doesn’t the KHP concentration have to be exactly 0.100M? Explain using complete sentences.
ollegr [7]

Answer:

The answer is in the explanation.

Explanation:

The KHP is an acid used as standard in titrations to find concentration of bases as NaOH.

The reaction that explain this use is:

KHP + NaOH → KNaP + H2O

<em>where 1 mole of KHP reacts per mole of NaOH</em>

That means, at equivalence point of a titration in which titrant is NaOH, the moles of KHP = Moles of NaOH added

With the moles of KHP = Moles of NaOH and the volume used by titrant we can find the molar concentration of NaOH.

The moles of KHP are obtained from the volume and the concentration as follows:

Volume(L)*Concentration (Molarity,M) = moles of KHP

If the concentration is more or less than 0.100M, the moles will be higher or lower. For that reason, we need to know the concentration of KHP but is not necessary to be 0.100M.

8 0
2 years ago
Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
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