How many milliliters of 0.250 M sodium hydroxide are needed to titrate 35.0 mL of 0.425 M hydrofluoric acid?
mLacid x Macid = mLbase x Mbase
To find the number of milliliters of 0.250 M sodium hydroxide needed to titrate 35.0 mL of 0.425 M hydrofluoric acid, you need to use the concept of molar ratio. The balanced chemical equation for this reaction is:
NaOH + HF → NaF + H2O
From the balanced chemical equation, we can see that the ratio between the moles of NaOH and HF is 1:1. This means that for every 1 mole of HF, we need 1 mole of NaOH.
First, let's calculate the moles of HF in 35.0 mL of 0.425 M hydrofluoric acid:
Moles of HF = volume (in liters) × concentration
= 35.0 mL × (1 L / 1000 mL) × 0.425 M
= 0.014875 mol
Since the molar ratio between NaOH and HF is 1:1, we know that we need the same number of moles of NaOH to neutralize the HF.
Now, let's calculate the volume of 0.250 M sodium hydroxide needed:
Volume of NaOH = moles of HF / concentration
= 0.014875 mol / 0.250 M
= 0.0595 L
Since the question asks for the volume in milliliters, we can convert the answer to milliliters:
Volume of NaOH = 0.0595 L × (1000 mL / 1 L)
= 59.5 mL
Therefore, 59.5 milliliters of 0.250 M sodium hydroxide are needed to titrate 35.0 mL of 0.425 M hydrofluoric acid.