a 350ml buffer solution is .150M in HF and .150M in NAF. What mass of naoh could this buffer neutralize before ph rises above 4.0?
So far I got .285714 from 10^-.544.
After that I tried to do .285714(x/.0525-x) but this doesn't worl. The answer is 1.2g. Help, please?
To find the mass of NaOH that the buffer solution can neutralize before the pH rises above 4.0, we need to consider the equilibrium reaction between HF and NaF:
HF + NaOH ⇌ NaF + H2O
From the given concentration of HF and NaF in the buffer solution, we can determine the initial moles of HF and NaF present. Let's start the calculations step by step:
Step 1: Calculate the initial moles of HF and NaF.
The initial moles of HF can be calculated using the following formula: moles = concentration × volume.
moles(HF) = 0.150 M × 0.350 L = 0.0525 moles
Similarly, the initial moles of NaF can be calculated as follows:
moles(NAF) = 0.150 M × 0.350 L = 0.0525 moles
Step 2: Identify the limiting reagent
In this case, HF and NaOH react in a 1:1 ratio (according to the balanced equation). Therefore, the limiting reagent will be the one with fewer moles, which is HF in this case.
Step 3: Determine the moles of NaOH that can react with the limiting reagent (HF)
Since HF and NaOH react in a 1:1 ratio, the moles of NaOH that can react with HF will be the same as the moles of HF present, which is 0.0525 moles.
Step 4: Calculate the mass of NaOH
To convert the moles of NaOH to grams, we need to multiply by its molar mass (40.00 g/mol).
mass of NaOH = moles of NaOH × molar mass of NaOH
mass of NaOH = 0.0525 moles × 40.00 g/mol = 2.10 g
Therefore, the mass of NaOH that this buffer solution can neutralize before the pH rises above 4.0 is 2.10 grams, not 1.2 grams. It seems there might have been an error in one of the previous steps of your calculation.