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What is the pH of a 0.5 M HF solution?
The pH of a 0.5 M HF solution is around 2.13. HF is a weak acid, so it partially dissociates in water to produce H+ ions, resulting in an acidic solution.
What is the pH level of 0.0005 M HF solution?
The pH of a 0.0005 M HF (hydrofluoric acid) solution can be calculated using the formula: pH = -log[H+]. First, determine the concentration of H+ ions produced by HF dissociation and then calculate the negative logarithm of that concentration to find the pH.
At a certain temperature a 0.25 M solution of HF has a pH of 2.3 What is the Ka of HF at this temperature?
To find the Ka of HF at this temperature, you can use the formula pH = pKa + log([A-]/[HA]). Since HF is a weak acid, [A-] is equal to [H+], so you can substitute [H+] for [A-] in the formula. Rearrange the formula to solve for pKa. Then convert pKa to Ka using the relationship: Ka = 10^(-pKa).
30 ml of hydrofluoric acid hf is titrated with a solution of 0.10 m koh if 55 ml of koh is required to reach the equivalence point what is the concentration of the acetic acid solution?
Since hydrofluoric acid (HF) and KOH react in a 1:1 ratio, the number of moles of KOH is equal to the number of moles of HF. Thus, the number of moles of KOH used in the titration is 55 ml * 0.10 M = 5.5 mmol. This is also the number of moles of HF present in 30 ml, so the concentration of the hydrofluoric acid solution is 5.5 mmol / 30 ml = 0.183 M.
If a 20.0g sample of HF is dissolved in water to give 200mL of solution what is the concentration of this solution?
concentration = molarity molarity= moles/ volume (in liters) H=1g F= 19g 1gH+ 19gF= 20gHF 20gHF x (1 mol HF/ 20gHF) = 1mol HF 2.0x10^2ml= 200ml x (1L/1000mL)= .200L Concentration= 1 mol HF/ .200L Concentration= 5M
What is the pH of a 0.5 M HF solution?
The pH of a 0.5 M HF solution is around 2.13. HF is a weak acid, so it partially dissociates in water to produce H+ ions, resulting in an acidic solution.
How many grams of HF are needed to make 0.50L of a 0.750 M solution?
The answer is 7,5g.
How many moles of HF Ka6.8104 must be present in 0.220dm3 to form a solution with a pH of 2.60?
An 0.010 M HF solution gives pH = 2.6, so 0.0022 molesare present in 0.22 L
What is the pH level of 0.0005 M HF solution?
The pH of a 0.0005 M HF (hydrofluoric acid) solution can be calculated using the formula: pH = -log[H+]. First, determine the concentration of H+ ions produced by HF dissociation and then calculate the negative logarithm of that concentration to find the pH.
At a certain temperature a 0.25 M solution of HF has a pH of 2.3 What is the Ka of HF at this temperature?
To find the Ka of HF at this temperature, you can use the formula pH = pKa + log([A-]/[HA]). Since HF is a weak acid, [A-] is equal to [H+], so you can substitute [H+] for [A-] in the formula. Rearrange the formula to solve for pKa. Then convert pKa to Ka using the relationship: Ka = 10^(-pKa).
How many grams of HF are needed to make 0.50 l of a 0.750 m solution?
To prepare a 0.750 m (molal) solution, you need to know the mass of the solvent (in kg) because molality is defined as moles of solute per kg of solvent. For a 0.50 L solution, assuming the solvent is water with a density of approximately 1 g/mL, you have 0.50 kg of water. Thus, for 0.750 m, you need 0.750 moles of HF. The molar mass of HF is about 20.01 g/mol, so you would need 0.750 moles × 20.01 g/mol = 15.01 grams of HF.
30 ml of hydrofluoric acid hf is titrated with a solution of 0.10 m koh if 55 ml of koh is required to reach the equivalence point what is the concentration of the acetic acid solution?
Since hydrofluoric acid (HF) and KOH react in a 1:1 ratio, the number of moles of KOH is equal to the number of moles of HF. Thus, the number of moles of KOH used in the titration is 55 ml * 0.10 M = 5.5 mmol. This is also the number of moles of HF present in 30 ml, so the concentration of the hydrofluoric acid solution is 5.5 mmol / 30 ml = 0.183 M.
Is ture or false the hydronium ion concentration of a solution that has a pH of 4.12 is 7.5910-9 M?
True, the hydronium ion concentration of a solution can be calculated using the formula [H3O+] = 10^(-pH). In this case, [H3O+] = 10^(-4.12) ≈ 7.59 x 10^(-5) M or 7.59 x 10^(-9) M.
If a 20.0g sample of HF is dissolved in water to give 200mL of solution what is the concentration of this solution?
concentration = molarity molarity= moles/ volume (in liters) H=1g F= 19g 1gH+ 19gF= 20gHF 20gHF x (1 mol HF/ 20gHF) = 1mol HF 2.0x10^2ml= 200ml x (1L/1000mL)= .200L Concentration= 1 mol HF/ .200L Concentration= 5M
A solution has OH- 3.5 x 10-6 Based on that what must be true about this solution?
This solution is basic (or alkaline) because it has a concentration of hydroxide ions (OH-) greater than 1 x 10^-7 M. It is likely that this solution has a pH greater than 7.
If the dipole moment of HF is 1.91 D then what is the dipole moment of HF in Cm?
1 Debye (D) is equal to 3.33564 x 10^-30 coulomb meters (Cm). Therefore, the dipole moment of HF in Cm would be 6.37669 x 10^-30 Cm (1.91 D * 3.33564 x 10^-30 Cm/D).
What is -10 equals -10 plus 7m?
It is a linear equation in the single variable, m. It has the solution m = 0
