Calculate the volume of 0.300M HNO3 needed to complete neutralize 25.0ml of 0.250M Sr(OH)2

2HNO3 + Sr(OH)2 ==> Sr(NO3)2 + 2H2O

mols Sr(OH)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Sr(OH)2 to mols HNO3.
Then M HNO3 = mols/L. You know mols and M HNO3, solve for L (and convert to mL if needed).

Well, well, trying to neutralize some chemistry here, huh? Don't you worry, I'll juggle some numbers for you!

To calculate the volume of 0.300M HNO3 needed to neutralize 25.0ml of 0.250M Sr(OH)2, we first need to find the molar ratio between the two compounds.

In the balanced chemical equation for this reaction, we have:

2HNO3 + Sr(OH)2 → Sr(NO3)2 + 2H2O

So, for every 2 moles of HNO3, we need 1 mole of Sr(OH)2.

Now, let's do some calculations!

First, let's find the moles of Sr(OH)2:

Moles of Sr(OH)2 = Volume in Liters x Molarity
= 0.0250 L x 0.250 mol/L
= 0.00625 mol

Since the molar ratio is 2HNO3:1Sr(OH)2, we can say that we need twice as many moles of HNO3 as Sr(OH)2:

Moles of HNO3 = 2 x Moles of Sr(OH)2
= 2 x 0.00625 mol
= 0.0125 mol

Now, let's find the volume of 0.300M HNO3 needed:

Volume of HNO3 = Moles of HNO3 / Molarity
= 0.0125 mol / 0.300 mol/L
= 0.0417 L

And voilà! The volume of 0.300M HNO3 needed to neutralize 25.0ml of 0.250M Sr(OH)2 is approximately 0.0417 liters (or 41.7 ml). I hope this helps you find your pH-ternal bliss!

To calculate the volume of 0.300M HNO3 needed to neutralize 25.0ml of 0.250M Sr(OH)2, we can use the stoichiometry of the balanced equation between these two compounds.

The balanced equation for the reaction between HNO3 and Sr(OH)2 is:

2 HNO3 + Sr(OH)2 -> 2 H2O + Sr(NO3)2

From this equation, we can see that it takes 2 moles of HNO3 to neutralize 1 mole of Sr(OH)2.

First, let's calculate the number of moles of Sr(OH)2 present in the solution:

moles of Sr(OH)2 = concentration (M) x volume (L)
= 0.250 M x 0.025 L
= 0.00625 moles

Since it takes 2 moles of HNO3 to neutralize 1 mole of Sr(OH)2, the number of moles of HNO3 needed can be calculated as:

moles of HNO3 needed = 2 x moles of Sr(OH)2
= 2 x 0.00625 moles
= 0.0125 moles

To find the volume of 0.300M HNO3 solution needed to contain 0.0125 moles of HNO3, we can rearrange the equation:

moles = concentration (M) x volume (L)

Rearranging for volume:

volume (L) = moles / concentration (M)

volume (L) = 0.0125 moles / 0.300 M
= 0.0417 L

Converting the volume from liters to milliliters:

volume (ml) = 0.0417 L x 1000
= 41.7 ml

Therefore, you would need approximately 41.7 ml of 0.300M HNO3 to neutralize 25.0 ml of 0.250M Sr(OH)2.

To calculate the volume of 0.300M HNO3 required to neutralize 25.0ml of 0.250M Sr(OH)2, we need to use the concept of stoichiometry and mole-to-mole ratios.

Step 1: Write the balanced chemical equation for the reaction between HNO3 and Sr(OH)2:
2HNO3 + Sr(OH)2 -> 2H2O + Sr(NO3)2

The equation shows that 2 moles of HNO3 react with 1 mole of Sr(OH)2 to produce 2 moles of water and 1 mole of Sr(NO3)2.

Step 2: Calculate the number of moles of Sr(OH)2:
moles = concentration (M) x volume (L)
moles of Sr(OH)2 = 0.250M x 0.0250L = 0.00625 moles

Step 3: Determine the mole-to-mole ratio between Sr(OH)2 and HNO3:
From the balanced equation, we see that 1 mole of Sr(OH)2 reacts with 2 moles of HNO3.

Step 4: Calculate the number of moles of HNO3 required:
moles of HNO3 = 0.00625 moles x (2 moles HNO3 / 1 mole Sr(OH)2)
moles of HNO3 = 0.0125 moles

Step 5: Calculate the volume of 0.300M HNO3 needed:
volume (L) = moles / concentration (M)
volume (L) = 0.0125 moles / 0.300M
volume (L) = 0.0417 L or 41.7 mL

Therefore, the volume of 0.300M HNO3 needed to neutralize 25.0ml of 0.250M Sr(OH)2 is approximately 41.7 mL.

41.6667mL