how many ml of 0.15 NaOH solution are required to neutralize 35.00 ml

mL x M = mL x M

5.25 To determine the number of milliliters of 0.15 M NaOH solution required to neutralize 35.00 ml of 0.22 moles of HCl, we can use the concept of stoichiometry.

Stoichiometry is a way of measuring the amounts of substances involved in chemical reactions. It involves using balanced chemical equations to determine the ratios of moles between reactants and products.

First, let's write the balanced chemical equation:

HCl + NaOH -> NaCl + H2O

From the equation, we can see that the molar ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, we need 1 mole of NaOH to neutralize it.

Given that the amount of HCl is 0.22 moles, we can conclude that we will need 0.22 moles of NaOH.

Next, let's calculate the volume of 0.15 M NaOH solution needed to contain 0.22 moles of NaOH. We can use the formula:

moles = concentration (M) x volume (L)

Rearranging the formula to solve for volume:

volume (L) = moles / concentration (M)

Plugging in the values:

volume (L) = 0.22 moles / 0.15 M

Now, we need to convert the volume from liters to milliliters since the given volume of HCl is in milliliters.

volume (L) = 0.22 moles / 0.15 M x 1000 ml/L

Simplifying:

volume (L) = 1466.67 ml / 0.15 M

volume (L) = 9777.78 ml

Therefore, approximately 9778 ml of the 0.15 M NaOH solution are required to neutralize 35.00 ml of 0.22 moles of HCl.

how many ml of 0.15 NaOH solution are required to neutralize 35.00 ml of 0.22 moles of HCL

mL x M = mL x M

5.25

To determine the number of milliliters of 0.15 M NaOH solution required to neutralize 35.00 ml of 0.22 moles of HCl, we can use the concept of stoichiometry.