The total acidity in water samples can be determined by neutralization with standard sodium hydroxide solution. What is the total concentration of hydrogen ion, H+, present in a water sample if 300. mL of the sample requires 7.5 mL of 2.8 10-3 M NaOH to be neutralized?
im confused as to what i am solving for?
is it H+ ions? can someone help me with this step by step?
It took 7.5 mL of 2.8 x 10^-3 M NaOH so that is M x L = moles HCl.
You have that many moles H^+ in the 300 mL sample of water; therefore, divide mole/L = M = concn H^+.
Sure, let's break it down step by step!
First, let's identify what we are trying to solve for. The question asks for the total concentration of hydrogen ions (H+) present in the water sample. So yes, we are solving for the H+ ions.
To do this, we can use the concept of stoichiometry. The balanced chemical equation for the neutralization between H+ ions and NaOH is:
H+ + OH- -> H2O
From this equation, we can see that 1 mole of H+ ions reacts with 1 mole of NaOH.
Now, let's calculate the number of moles of NaOH used in the neutralization. We know that 7.5 mL of 2.8 x 10^-3 M NaOH is used. To do this, we'll use the formula:
moles = concentration x volume
moles of NaOH = (2.8 x 10^-3 M) x (7.5 mL / 1000)
moles of NaOH = 2.1 x 10^-5 moles
Since 1 mole of NaOH reacts with 1 mole of H+ ions, we can say that the number of moles of H+ ions present in the water sample is also 2.1 x 10^-5 moles.
Finally, let's calculate the total concentration of H+ ions in the water sample. We know that 300 mL of the sample was used in the titration. So, the concentration can be calculated using the formula:
concentration = moles / volume
concentration of H+ ions = (2.1 x 10^-5 moles) / (300 mL / 1000)
concentration of H+ ions = 7 x 10^-5 M
Therefore, the total concentration of hydrogen ions in the water sample is 7 x 10^-5 M.
I hope this helps! Let me know if you have any other questions.
Yes, you are correct. You are solving for the concentration of hydrogen ions (H+) in the water sample.
To determine the concentration of H+ ions, you can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction. The balanced equation for the neutralization reaction between H+ ions and NaOH is:
H+ (aq) + OH- (aq) → H2O (l)
From the balanced equation, you can see that for every 1 mole of H+ ions, 1 mole of NaOH is required for neutralization. Therefore, the number of moles of H+ ions can be calculated using the following formula:
Number of moles of H+ ions = Concentration of NaOH (M) × Volume of NaOH (L)
Now let's calculate the number of moles of H+ ions.
Concentration of NaOH = 2.8 × 10^-3 M
Volume of NaOH = 7.5 mL = 7.5 × 10^-3 L (converted to liters by dividing by 1000)
Number of moles of H+ ions = (2.8 × 10^-3 M) × (7.5 × 10^-3 L)
= 2.1 × 10^-5 moles
Since the neutralization reaction is 1:1 between H+ and NaOH, the number of moles of H+ ions is equal to the number of moles of NaOH used.
Since the volume of the water sample is not given, we cannot directly calculate the concentration of H+ ions in the water sample. However, we can use the dilution formula to determine the original concentration of H+ ions in the water sample.
Dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration of H+ ions in the water sample
V1 = initial volume of the water sample (unknown)
C2 = final concentration of H+ ions after neutralization (which is equal to the molarity of NaOH used in the neutralization)
V2 = final volume of the NaOH solution used
Using the dilution formula, you can rearrange it to solve for C1:
C1 = (C2 × V2) / V1
Since C2 = 2.8 × 10^-3 M (the molarity of NaOH used) and V2 = 7.5 × 10^-3 L (the volume of NaOH used), we can substitute these values into the equation:
C1 = (2.8 × 10^-3 M × 7.5 × 10^-3 L) / V1
Now, you need to insert the volume of the water sample, 300 mL (converted to liters by dividing by 1000), into the equation:
C1 = (2.8 × 10^-3 M × 7.5 × 10^-3 L) / 300 × 10^-3 L
= 7 × 10^-5 M
Therefore, the total concentration of hydrogen ions (H+) present in the water sample is 7 × 10^-5 M.
Yes, you are correct. In this problem, you are solving for the total concentration of hydrogen ions, H+, present in the water sample.
To solve this problem, you can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between H+ ions and sodium hydroxide (NaOH):
H+ + OH- → H2O
Here's a step-by-step breakdown of how to solve the problem:
Step 1: Convert the given volume of the water sample to liters.
The given volume is 300. mL. Since the standard unit for volume in the International System of Units (SI) is liters, you need to convert 300. mL to liters. Recall that there are 1000 mL in one liter, so:
300. mL ÷ 1000 mL/L = 0.3 L
Step 2: Determine the number of moles of NaOH used.
To find the number of moles, you can use the equation:
moles = concentration (M) × volume (L)
The given concentration of NaOH is 2.8 × 10^-3 M, and the volume of NaOH used is 7.5 mL. Convert the volume to liters:
7.5 mL ÷ 1000 mL/L = 0.0075 L
Now, calculate the number of moles of NaOH:
moles = (2.8 × 10^-3 M) × (0.0075 L) = 2.1 × 10^-5 moles
Step 3: Determine the number of moles of H+ ions in the water sample.
From the balanced chemical equation, you can see that one mole of NaOH reacts with one mole of H+ ions. Therefore, the number of moles of H+ ions is equal to the number of moles of NaOH used.
Number of moles of H+ ions = 2.1 × 10^-5 moles
Step 4: Calculate the total concentration of H+ ions in the water sample.
The concentration of H+ ions is expressed in moles per liter (M). Since you have the number of moles of H+ ions and the volume of the water sample in liters (0.3 L), divide the number of moles by the volume:
Total concentration of H+ ions = (2.1 × 10^-5 moles) / (0.3 L) = 7.0 × 10^-5 M
So, the total concentration of hydrogen ions, H+, in the water sample is 7.0 × 10^-5 M.