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ale4655 [162]
2 years ago
9

Which statement best describes the pH of pure water? It is neutral because the concentration of hydronium ions equals that of hy

droxide ions. It is neutral because the pure liquid contains neither hydronium ions nor hydroxide ions. It is acidic because it has a hydronium ion concentration of mc027-1. Jpg. It is basic because it has a hydroxide ion concentration of mc027-2. Jpg.
Chemistry
2 answers:
stira [4]2 years ago
4 0

The pH of pure water has been best described as neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.

pH has been described as the measurement of hydrogen ions in a solution. The pH has been measured on a scale of 1-14. pH 7 has been the neutral pH.

The higher hydronium ion concentration tends to move the pH from 7 towards 1. The higher hydroxide ion concentration tends to move the pH above 7.

The neutral pH has been neither acidic nor basic with the equal constituents of hydronium and hydroxide ion in the solution.

Thus, the pH of pure water has been 7. It has neutral pH with equal hydronium and hydroxide ions. Thus, option A is correct.

For more information about the pH of the solution, refer to the link:

brainly.com/question/4975103

GalinKa [24]2 years ago
3 0

Answer:

Water is neutral because there is an equal concentration of hydronium ions and hydroxide ions.

Explanation:

Water autoionizes into equal parts hydronium and hydroxide ions, resulting in a neutral pH. This is because during the autoionization reaction, two water molecules yield one hydronium and one hydroxide.

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An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
RSB [31]

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

Best regards!

7 0
3 years ago
A solution is prepared by dissolving 38.6 g sucrose (C12H22O11) in 495 g of water. Determine the mole fraction of sucrose if the
stiks02 [169]

Answer:

4.09×10⁻³ is the mole fraction of sucrose

Explanation:

Mole fraction = Moles of solute or solvent/ Total moles

Let's convert the mass to moles (mass / molar mass)

38.6 g / 342 g/m = 0.113 moles of sucrose

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Total moles = 0.113 m + 27.5 m = 27.0613 moles

Mole fraction of sucrose = Moles of sucrose / Total moles

0.113 m / 27.0613 moles = 4.09×10⁻³

8 0
3 years ago
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