How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of 0.389 M
HNO3? Show your working by writing balanced equation.
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How many milliliters of 1.25 mol / dm3 LiOH must be added to neutralize 34.7 ml of 0.389 mol / dm3 HNO3?
To find the amount of LiOH needed to neutralize HNO3, we first need to write a balanced equation for the reaction between LiOH and HNO3.
The balanced equation for the reaction between LiOH and HNO3 is:
2 LiOH + HNO3 -> Li2O + 2 H2O + NO
From the balanced equation, we can see that the stoichiometric ratio between LiOH and HNO3 is 2:1. This means that 2 moles of LiOH react with 1 mole of HNO3.
Now, let's calculate the number of moles of HNO3 using the given concentration and volume:
moles of HNO3 = concentration of HNO3 * volume of HNO3
= 0.389 M * 0.0347 L
= 0.0134683 moles
Since the stoichiometric ratio of LiOH to HNO3 is 2:1, we can calculate the number of moles of LiOH needed:
moles of LiOH = 2 * moles of HNO3
= 2 * 0.0134683 moles
= 0.0269366 moles
Finally, we can find the volume of 1.25 M LiOH solution needed using its concentration:
volume of LiOH = moles of LiOH / concentration of LiOH
= 0.0269366 moles / 1.25 M
= 0.021549 moles
Since the concentration is given in M (moles per liter), we need to convert moles to milliliters by multiplying by 1000:
volume of LiOH = 0.021549 moles * 1000 mL
= 21.549 mL
Therefore, you would need to add approximately 21.549 mL of 1.25 M LiOH to neutralize 34.7 mL of 0.389 M HNO3.
LiOH + HNO3 ==> LiNO3 + H2O
mLacid + M acid = mL base + M base.
Substitute and solve for the one unknown.