- log(0.000626 M H2SO4)
= 3.2 pH
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The pH of a 0.1 M solution of sulfuric acid (H2SO4) is around 1. Note that the pH of sulfuric acid solutions will vary depending on concentration.
To prepare a 0.005 M solution of H2SO4 with pH 4, you can first calculate the concentration of H+ ions needed to achieve a pH of 4. Then, use the dissociation of H2SO4 to determine the amount of H2SO4 needed to provide that concentration of H+ ions. Finally, dilute the calculated amount of H2SO4 with water to reach the desired volume of the solution.
The sulfuric acid (H2SO4) will dissociate into H+ and SO4^2-. Since each molecule of sulfuric acid produces 2 H+ ions, the concentration of H+ ions will be 0.02M. The pH of this solution can be calculated as pH = -log[H+], so pH = -log(0.02) ≈ 1.7.
The first solution is more concentrated because it contains 6 moles of H2SO4 per one liter of solution. The second solution is less concentrated because it contains 0.1 moles of H2SO4 in one liter. In equal amounts of each example, the first would have more H2SO4.
To dilute the 3.5 M H2SO4 solution to 2 M, you need to add water. Use the formula M1V1 = M2V2, where M represents molarity and V represent volume. For this situation, you'll end up adding 75 ml of water to the initial 75 ml of 3.5 M H2SO4 solution to achieve a final 2 M concentration.
The pH of a 0.1 M solution of sulfuric acid (H2SO4) is around 1. Note that the pH of sulfuric acid solutions will vary depending on concentration.
To prepare a 0.005 M solution of H2SO4 with pH 4, you can first calculate the concentration of H+ ions needed to achieve a pH of 4. Then, use the dissociation of H2SO4 to determine the amount of H2SO4 needed to provide that concentration of H+ ions. Finally, dilute the calculated amount of H2SO4 with water to reach the desired volume of the solution.
The sulfuric acid (H2SO4) will dissociate into H+ and SO4^2-. Since each molecule of sulfuric acid produces 2 H+ ions, the concentration of H+ ions will be 0.02M. The pH of this solution can be calculated as pH = -log[H+], so pH = -log(0.02) ≈ 1.7.
The first solution is more concentrated because it contains 6 moles of H2SO4 per one liter of solution. The second solution is less concentrated because it contains 0.1 moles of H2SO4 in one liter. In equal amounts of each example, the first would have more H2SO4.
To dilute the 3.5 M H2SO4 solution to 2 M, you need to add water. Use the formula M1V1 = M2V2, where M represents molarity and V represent volume. For this situation, you'll end up adding 75 ml of water to the initial 75 ml of 3.5 M H2SO4 solution to achieve a final 2 M concentration.
0.08 n
To find the volume of 6.40 M H2SO4 needed to prepare a solution that is 0.700 M, you can use the formula: M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. Rearranging the formula, V1 = (M2*V2) / M1 gives you the volume: V1 = (0.700 * 455.0) / 6.40 = 49.53 ml. You would need 49.53 ml of the 6.40 M H2SO4 solution to prepare 455.0 ml of a 0.700 M H2SO4 solution.
The pH of a solution containing 0.1 M of HC2H3O2 is around 2.88.
its PH is 3
The pH of a solution can be calculated using the formula: pH = -log[OH-]. Therefore, for a solution with [OH-] concentration of 10-12 M, the pH would be 12.
Remember M1V1=M2V2, where M is molarity and V is volume. M1/M2=V2/V1, 10/1=v2/v1, For diluting the acid, we can add acid to water. So, assuming that 10M H2SO4 is having 1ml of water, we should add 1M of H2So4 to 10ml of water.
The approximate pH of a 1 M solution of Na2CO3 is around 11-12, making it basic.