In procedure B2, you will titrate a saturated solution to find out how much of the anion dissolved. Assume you we're able to dissolve about 0.12g of potassium hydrogen tartrate in 25.0 mL of water. For the titration, you will use 10.00mL of this sample. How many mL of 0.0500M NaOH would required to bring the titration to its end point? Show your work.
Someone help idk where to start
mols potassium hydrogen tartrate = grams/molar mass = 0.12/molar mass = ?
You took 10/25 of this for titration; therefore, mols titrated = mols initially x 10/25 = ?
M NaOH = mols tartrate titrated/L NaOH.
You know M NaOH and mols titrated, solve for L NaOH and convert to mL.
Thank you so much
To solve this problem, we need to understand the concept of titration and how it is used to determine the concentration of a solution.
Titration is a process in which a solution of known concentration, called the titrant, is added to a solution of unknown concentration until a reaction between the two is complete. This reaction is often indicated by an observable change, known as the endpoint of the titration. By measuring the volume of the titrant required to reach the endpoint, we can calculate the concentration of the unknown solution.
Now let's break down the given information and solve the problem step by step:
1. We are given that a saturated solution of potassium hydrogen tartrate (KHT) was prepared with 0.12g of the compound dissolved in 25.0 mL of water.
2. From this saturated solution, we will use 10.00 mL for the titration.
3. The titrant is a 0.0500 M solution of sodium hydroxide (NaOH).
To find out how many milliliters of NaOH are needed to reach the endpoint, we need to set up an equation based on the balanced chemical equation for the reaction between KHT and NaOH:
KHT + NaOH -> KNaT + H2O
From the equation, we can see that the ratio between the number of moles of KHT and NaOH is 1:1. Therefore, the number of moles of NaOH required to neutralize the KHT is equal to the number of moles of KHT in the sample.
Now let's calculate the number of moles of KHT in our sample:
- The molar mass of KHT (also called tartaric acid) is approximately 150.086 g/mol. So, the number of moles of KHT in 0.12 g can be calculated as follows:
moles of KHT = mass of KHT / molar mass of KHT
= 0.12 g / 150.086 g/mol
≈ 0.000798 mol
Since the ratio between KHT and NaOH is 1:1, we know that the same number of moles of NaOH is required to reach the endpoint.
Now let's calculate the volume of NaOH required using the given molarity and the number of moles of NaOH:
- The molarity (M) of NaOH is given as 0.0500 M.
moles of NaOH = moles of KHT
volume of NaOH (in L) = moles of NaOH / molarity of NaOH
= 0.000798 mol / 0.0500 mol/L
= 0.01596 L
To convert the volume from liters to milliliters, we can multiply it by 1000:
volume of NaOH (in mL) ≈ 0.01596 L * 1000 mL/L ≈ 15.96 mL
Rounding to the nearest hundredth, we find that approximately 15.96 mL of 0.0500 M NaOH are required to bring the titration to its endpoint.
Remember, it's important to double-check all calculations and consider significant figures throughout the process.