You prepare a solution by dissolving 5.07 grams of K2HPO43•H2O in about 50 mL of

Question

You prepare a solution by dissolving 5.07 grams of K2HPO43•H2O in about 50 mL of water, which you transfer to a 250.0 mL volumetric flask. What volume of which solution should you add so that the final mixture is at pH 7.50 after filling the flask to the mark with deionized water?

Available solids benzoic acid (C6H5COOH), sodium chlorite (NaClO2), potassium hydrogen phosphate trihydrate
Available solutions (K2HPO4•3H2O) 0.1973 M HCl, 0.1309 M NaOH, 2.142 M ammonia
You prepare a solution by dissolving 5.07 grams of K2HPO43•H2O in about 50 mL of water, which you transfer to a 250.0 mL volumetric flask. What volume of which solution should you add so that the final mixture is at pH 7.50 after filling the flask to the mark with deionized water?

Available solids benzoic acid (C6H5COOH), sodium chlorite (NaClO2), potassium hydrogen phosphate trihydrate
Available solutions (K2HPO4•3H2O) 0.1973 M HCl, 0.1309 M NaOH, 2.142 M ammonia

Available solids benzoic acid (C6H5COOH), sodium chlorite (NaClO2), potassium hydrogen phosphate trihydrate
Available solutions (K2HPO4•3H2O) 0.1973 M HCl, 0.1309 M NaOH, 2.142 M ammonia
benzoic acid (C6H5COOH), sodium chlorite (NaClO2), potassium hydrogen phosphate trihydrate
Available solutions (K2HPO4•3H2O) 0.1973 M HCl, 0.1309 M NaOH, 2.142 M ammonia

Explanation / Answer

Molar mass of K2HPO43(H2O) is 816.1678 g/mol

You prepare a solution by dissolving 5.07 grams of K2HPO43•H2O

moles = 5.07/816.16 =0.0062

molarity = 0.0062/0.050 =0.124 M

pH = 7.5 is the desired

so, we have a stock solution of 0.124M K2HPO4

now, diluting it to 250 ml will make 0.0248M

x volume of 0.1973 M HCl and 0.248M KH2PO4 has to be used to get pH =7.5 in 250 ml solution

The reaction between HCl and K2HPO4 yields KH2PO4 and KCl. KCl does not affect the pH much

now, let's calculate volume of HCl

pH = pka + log [base]/[acid]

7.5 = 7.20 + log [base]/[acid]

[base] =0.248

concentration of KH2PO4 required is 0.022

so, let's check the volume of HCl that should be give 0.022M KH2PO4

It is 0.022 M

volume X 0.1973 = 0.022X 250 ml = 27.86ml of given HCL solution

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