How do you set up this problem? Calculate the volume , in ml of a 0.150 M NaOH soulution needed to neutralize 25.0ml of a 0.288 M HCl solution?
Since each mole of HCl neutralizes 1 mole of NaOH,
.025L * .288M = 0.0072 moles HCl
So, for the NaOH,
0.0072 moles / 0.150M = .048L or 48 ml
To set up this problem, you need to start by writing the balanced chemical equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
From the equation, you can see that one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water (H2O).
Next, you need to determine the number of moles of HCl in the 25.0 mL solution. To do this, you can use the formula:
moles = concentration (M) × volume (L)
Convert the volume from mL to L:
25.0 mL = 25.0 mL × (1 L / 1000 mL) = 0.025 L
Now, substitute the values into the formula:
moles of HCl = 0.288 M × 0.025 L = 0.0072 moles
Since the NaOH and HCl react in a 1:1 ratio, the number of moles of NaOH required to neutralize the HCl is also 0.0072 moles.
To calculate the volume of the NaOH solution needed, you can again use the formula:
volume (L) = moles / concentration (M)
Substitute the values into the formula:
volume of NaOH = 0.0072 moles / 0.150 M = 0.048 L
Finally, convert the volume back to mL:
volume of NaOH = 0.048 L × (1000 mL / 1 L) = 48 mL
Therefore, the volume of the 0.150 M NaOH solution needed to neutralize 25.0 mL of the 0.288 M HCl solution is 48 mL.