Asked by K
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
Someone help idk where to start
Answers
Answered by
DrBob222
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.
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.
Answered by
K
Thank you so much
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