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Lana71 [14]
1 year ago
13

More strong base is added until the equivalence point is reached. what is the ph of this solution at the equivalence point if th

e total volume is 57. 0 mlml ?
Chemistry
1 answer:
Vlad [161]1 year ago
7 0

A more strong base is added until the equivalence point is reached. The ph of this solution at the equivalence point if the total volume is 57. 0 mill is 9.8

The potential of Hydrogen is what pH is formally known as. The negative logarithm of the concentration of H+ ions is known as pH. Thus, the definition of pH as the amount of hydrogen is provided. The hydrogen ion concentration in a solution is described by the pH scale, which also serves as a gauge for the solution's acidity or basicity.

The pH scale determines how acidic or basic water is. The range is 0 to 14, with 7 representing neutrality. Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7. In reality, pH is a measurement of the proportion of free hydrogen and hydroxyl ions in water.

To learn more about pH please visit -
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If you dissolve 50.0 grams of potassium bromide in 100.0g of water what is the mass percent of the resultant solution?
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Mass percent= grams solute/ grams of solution x 100

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What is the number of atoms per unit cell for each metal?<br> (c) Silver, Ag
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Silver (Ag) is the number of atoms per unit cell for each metal. Silver has a face-centred cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.

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5 0
1 year ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci
Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

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