What volume of 0.45 M LiOH is needed to neutralize 60.0 mL of 0.15 M Hl?

To determine the volume of 0.45 M LiOH (lithium hydroxide) needed to neutralize 60.0 mL of 0.15 M Hl (hydroiodic acid), we can perform a titration calculation using the concept of moles and the balanced chemical equation of the neutralization reaction. The balanced equation for the reaction between LiOH and HI is:

LiOH + HI → LiI + H2O

From the equation, we see that LiOH reacts with HI in a 1:1 molar ratio. This means that one mole of LiOH will neutralize one mole of HI.

First, we need to calculate the number of moles of HI present in 60.0 mL of the 0.15 M solution:

moles of HI = molarity of HI * volume of HI in liters moles of HI = 0.15 mol/L * 60.0 mL * (1 L / 1000 mL) moles of HI = 0.15 * 0.0600 moles of HI = 0.009 moles

Since the molar ratio of LiOH to HI is 1:1, the moles of LiOH needed will also be 0.009 moles.

Next, we calculate the volume of the 0.45 M LiOH solution that contains 0.009 moles of LiOH:

volume of LiOH = moles of LiOH / molarity of LiOH volume of LiOH = 0.009 moles / 0.45 mol/L volume of LiOH = 0.020 liters

To express the volume in milliliters, we convert liters to milliliters:

volume of LiOH = 0.020 L * 1000 mL/L volume of LiOH = 20.0 mL

So, 20.0 mL of 0.45 M LiOH is needed to neutralize 60.0 mL of 0.15 M HI.