A 250 ml sample of saturated a g o h solution was titrated with h c l , and the endpoint was reached after 2. 60 ml of 0. 0136 m h c l was dispensed. Based on this titration, what is the k s p of a g o h <u>. Ksp=1.9×10⁻⁸</u>
<h3>What is titration?</h3>
Titration is a typical laboratory technique for quantitative chemical analysis used to calculate the concentration of a specified analyte. It is also referred to as titrimetry and volumetric analysis (a substance to be analyzed). A standard solution with a known concentration and volume is prepared as the reagent, also known as the titrant or titrator. To ascertain the concentration of the analyte, the titrant reacts with an analyte solution (also known as the titrand). The titration volume is the amount of titrant that interacted with the analyte.
A typical titration starts with a beaker or Erlenmeyer flask being placed below a calibrated burette or chemical pipetting syringe that contains the titrant and a little amount of the indicator (such as phenolphthalein).
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I think the answer would be Ionic sodium phosphate (Na3PO4) because it has the greatest boiling point elevation.
Answer:
0.0295M
Explanation:
As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.
You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.
That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.
And the concentration of SCN⁻ is:
0.0800M / 2.714 =
<h3>0.0295M </h3>
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