Calculate the volume of 2.00 M solution of KOH that is required to neutralize 30.0 mL of 1.00 M solution of HNO3.
volumeacid*Macid*Heqa=volumeb*Mbase*Heqb
30ml*1M*1=Vb*2M*1
solve for Vb
After that part im lost i know the formula bit it didn't make sense
I got this , thank you
To calculate the volume of the 2.00 M solution of KOH required to neutralize 30.0 mL of 1.00 M solution of HNO3, we need to use the concept of stoichiometry.
The balanced chemical equation for the neutralization reaction between KOH and HNO3 is:
KOH + HNO3 → KNO3 + H2O
From the balanced equation, we can see that the molar ratio between KOH and HNO3 is 1:1. This means that 1 mole of KOH reacts with 1 mole of HNO3.
Given that the volume of the HNO3 solution is 30.0 mL (which is equal to 0.030 L) and its concentration is 1.00 M, we can calculate the number of moles of HNO3 using the formula:
moles of solute = concentration × volume
moles of HNO3 = 1.00 M × 0.030 L
Now that we know the number of moles of HNO3, we can use the stoichiometric ratio to determine the number of moles of KOH required. Since the ratio is 1:1, the number of moles of KOH will be the same as the number of moles of HNO3.
Finally, to find the volume of the 2.00 M solution of KOH, we can use the formula:
volume = moles of solute / concentration
volume of KOH = moles of KOH / 2.00 M
By substituting the value of moles of KOH, which is the same as the moles of HNO3 calculated earlier, we can find the volume of the KOH solution required to neutralize the HNO3 solution.