Determine the pH of the following solutions.
1) 0.600 M HCl
2) 0.052 moles of H2SO4 dissolved in 4.2 L of water.
3)5.6 grams HNO3 dissolved in 450 ml of water.
4) 2.5 grams of NaOH dissolved in 500 mL of water.
5) adding 2 grams NaH dissolved in 0.56 L of water.
6) 20 mL of 0.27 M HNO3 is diluted to 500 mL.
Tyra, you really need to help by answering the questions we ask. I need to know WHAT you don't understand about a problem.
1. For strong acids like HCl which ionize 100%, then pH = -log HCl)
3. Convert 5.6 g to mols. mols = grams/molar mass. Then the definition of molarity is M = mols/L. You know mols from the conversion you made from 5.6 grams and you know it is in 0.450 L.
4. is the same procedure as 3.
6. For dilutions use the dilution formula of
mL1 x M1 = mL2 x M2
20 x 0.27 = 500 x ?
Solve for ?
0.0108
To determine the pH of a solution, you need to know the concentration of H+ ions in the solution. One way to do this is by using the equation pH = -log[H+], where [H+] is the concentration of H+ ions in moles per liter.
Let's go through each solution step by step:
1) 0.600 M HCl:
HCl is a strong acid that dissociates completely in water. Therefore, the concentration of H+ ions in this solution is equal to the concentration of HCl. So, the pH is given by pH = -log(0.600) = 0.22.
2) 0.052 moles of H2SO4 dissolved in 4.2 L of water:
H2SO4 is also a strong acid and dissociates completely in water. Since we have the number of moles of H2SO4, we can calculate the molarity (M) by dividing moles by the volume in liters: M = 0.052 moles / 4.2 L = 0.012 M. Therefore, the concentration of H+ ions is 0.012 M, and the pH is pH = -log(0.012) ≈ 1.92.
3) 5.6 grams HNO3 dissolved in 450 ml of water:
First, we need to convert grams to moles. The molar mass of HNO3 is 63.01 g/mol. We can calculate the number of moles of HNO3 using the formula: moles = grams / molar mass = 5.6 g / 63.01 g/mol ≈ 0.089 moles. Now we can calculate the molarity by dividing the number of moles by the volume in liters: M = 0.089 moles / 0.450 L = 0.198 M. Therefore, the concentration of H+ ions is 0.198 M, and the pH is pH = -log(0.198) ≈ 0.70.
4) 2.5 grams of NaOH dissolved in 500 mL of water:
Similar to the previous example, we need to convert grams to moles. The molar mass of NaOH is 39.997 g/mol. We can calculate the number of moles of NaOH using the formula: moles = grams / molar mass = 2.5 g / 39.997 g/mol ≈ 0.063 moles. Now we can calculate the molarity by dividing the number of moles by the volume in liters: M = 0.063 moles / 0.500 L = 0.126 M. However, NaOH is a strong base, so we need to consider the OH- ions. In this case, the concentration of OH- ions would be 0.126 M. To find the pOH, we use pOH = -log[OH-] = -log(0.126) ≈ 0.90. Finally, since pH + pOH = 14 (at 25 °C), the pH would be pH = 14 - pOH = 14 - 0.90 ≈ 13.10.
5) Adding 2 grams NaH dissolved in 0.56 L of water:
Similar to previous examples, we need to convert grams to moles. The molar mass of NaH is 22.99 g/mol. We can calculate the number of moles of NaH using the formula: moles = grams / molar mass = 2 g / 22.99 g/mol ≈ 0.087 moles. Now we can calculate the molarity by dividing the number of moles by the volume in liters: M = 0.087 moles / 0.56 L = 0.155 M. However, NaH is a strong base that dissociates completely, leaving only the OH- ions. So, the concentration of OH- ions is 0.155 M. Using the same logic as in the previous example, we can find the pOH = -log(0.155) ≈ 0.81. Therefore, the pH = 14 - pOH = 14 - 0.81 ≈ 13.19.
6) 20 mL of 0.27 M HNO3 is diluted to 500 mL:
To determine the concentration after dilution, we can use the formula: M1V1 = M2V2, where M1 and V1 are the initial concentration and volume, and M2 and V2 are the final concentration and volume. Plugging in the values, we have (0.27 M)(20 mL) = M2(500 mL). Solving for M2, we get M2 = (0.27 M)(20 mL) / 500 mL ≈ 0.0108 M. Therefore, the concentration of H+ ions is 0.0108 M, and the pH is pH = -log(0.0108) ≈ 1.97.
Remember, these calculations rely on the assumption that the substances are fully dissociated in water, like strong acids and bases.