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does kh2po4 react with sodium lauryl sulphate and can we use nah2po4 insted of kh2po4 in prepration of 0.2 m phosphate buffer ?
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if 4 volumes of 0.1M monobasic potassium phosphate, KH2PO4, are mixed with 2 volume of 0.1 M dibasic sodium phosphate, Na2HPO4,
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Use pH = pKa + log[(base)/(acid)] KH2PO4 is the acid. Na2HPO4 is the base. For b part: Write the
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What is pH buffer of 0.1 mol L-1 Na2HPO4/O.15 mol L-1 KH2PO4?
KH2PO4? Given Ka(H2PO4-) = 6.2 x 10-8). I would like to know the
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pH = pKa + log [(base)/(acid)] pH = 7.2 + log (0.1/0.15) [H = 7.02. Check my work.
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This question pertaining to buffer systems!
How would you prepare 10.0mL of 0.0100M phosphate buffer, pH = 7.4 0, from stock
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pH = pK2 + log b/a 7.4 = 7.2 + log b/a log b/a = 1.58. You're right, I typed in the wrong numbers in
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If you mix 30 mL of 0.05 M K2HPO4 with 24 mL of 0.05 M KH2PO4, what % of the PO4 buffer is in the K2HPO4 form?
I'm just not sure
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Yes, you are correct that both K2HPO4 and KH2PO4 can act as buffers. However, in this question, we
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Suppose you have to prepare 1.00 L of the solution ( KH2PO4 and Na2HPO4 pH=7.31) and that this solution must be isotonic with
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To determine the mass of KH2PO4 needed to prepare the isotonic solution, we need to understand a few
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A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl
Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3 ;
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Use the Henderson-Hasselbalch equation.
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using the following salts listed below, calculate the moles of acid and conjugate base needed to make 100 ml of 0.50 M, ph
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pH = pKa + log(base)/(acid) You want 100 mL x 0.5M = 0.05 mols 6.5 = 6.64 + log (B/A) Solve for B/A,
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what mass of sodium sulphate is produced if 49g of hydrogen sulphate react with 80g of NAOH?
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2NaOH + H2SO4 --> Na2SO4 + 2H2O This is a limiting reagent (LR) problem. You know that when an
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40grams of sodium hydroxide was reacted with sulphuric acid to give 35grams sodium sulphate.calculate the percentage yield of
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Here is an example. http://www.jiskha.com/science/chemistry/stoichiometry.html
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complete KH2PO4(aq)+KOH(aq)
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KH2PO4 + KOH ==> K3PO4 + H2O You balance it.
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