If you dissolve 30.0 g of HF in 100mL of water, what is the concentration of H+ ions in this solution? (Can you figure out what the H+ concentration would be using the Ka value?) Ka=6.3*10^-4
Yes, the concn HF is moles/L. You get moles from g HF/molar mass and the volume is 0.1 L. Then use Ka for HF, set up an ICE chart and solve for H^+.
To find the concentration of H+ ions, we can use the Ka value of HF and the initial concentration of HF in the solution. The equation for the dissociation of HF in water is:
HF (aq) ⇌ H+ (aq) + F- (aq)
The Ka expression for this reaction is:
Ka = [H+][F-] / [HF]
Since the initial concentration of HF is given as 30.0 g in 100 mL of water, we need to convert this mass concentration to molar concentration before we can use it in the Ka expression.
First, calculate the molar mass of HF. The molar mass of hydrogen (H) is approximately 1 g/mol and the molar mass of fluorine (F) is approximately 19 g/mol.
Molar mass of HF = 1 g/mol + 19 g/mol = 20 g/mol
Next, convert the mass of HF to moles:
moles of HF = mass of HF / molar mass of HF
moles of HF = 30.0 g / 20 g/mol
moles of HF = 1.5 mol
Now, we need to convert the volume of the solution to liters:
volume of solution = 100 mL = 100/1000 L = 0.1 L
Using the balanced equation, we can see that the moles of H+ ions produced will be the same as the moles of HF initially present. Therefore, the concentration of H+ ions is equal to the initial concentration of HF in moles per liter.
concentration of H+ ions = moles of HF / volume of solution
concentration of H+ ions = 1.5 mol / 0.1 L
concentration of H+ ions = 15 mol/L
Therefore, the concentration of H+ ions in the solution is 15 mol/L.
Please note that this calculation assumes 100% dissociation of HF in water, which may not be the case in reality.