how many ml of 6M HCl solution are required to bring the pH of 10 ml of a 4M NaOH solution to 2.0
To solve this problem, we need to use the concept of neutralization reaction between an acid and a base. In this case, we have a 4M NaOH solution and we want to bring its pH down to 2.0 by adding a 6M HCl solution.
Since NaOH is a strong base and HCl is a strong acid, they will react in a 1:1 ratio according to the equation:
NaOH + HCl -> NaCl + H2O
From the equation, we can deduce that for every mole of NaOH, we need 1 mole of HCl to neutralize it completely.
First, let's calculate the number of moles of NaOH in the 10 ml of 4M NaOH solution.
The molarity (M) is defined as moles of solute per liter of solution. Here, we have a 4M NaOH solution, which means there are 4 moles of NaOH in 1 liter of solution.
So, the number of moles of NaOH in 10 ml = (4 moles/L) x (10 ml/1000 ml) = 0.04 moles
Since we need an equal number of moles of HCl to neutralize the NaOH, we need 0.04 moles of HCl.
Now, let's calculate the volume (in ml) of the 6M HCl solution required to have 0.04 moles of HCl.
The molarity is defined as moles of solute per liter of solution. Here, we have a 6M HCl solution.
So, the volume of HCl solution = (0.04 moles) / (6 moles/L) = 0.00667 L = 6.67 ml
Therefore, you would require approximately 6.67 ml of 6M HCl solution to bring the pH of 10 ml of a 4M NaOH solution to 2.0.