Question

Calculate the volume, in milliliters, of a 0.390 M KOH solution that should be added to 5.500 g of HEPES (MW = 238.306 g/mol, PK, = 7.56) to give a pH of 7.58. KOH volume = 17

Answer 1

Let, NmL of 00390M KOH solution was added. Hoes, NmL of 0.390 M KOH = NmKX 0.39 0 mod it = 0.390 Xw Anmol of KOH = 0.00039 0 V mol of Koht 5.500 g of HERES = 5.500 4 = 0.023079569 mol of MERES i 238.306 Amofos Logleally here HEPES will be in ences to form acidie buffel. Now, I mol of Koh seacts with I mol of heres to form 1 mol Hof, K-selt of HEPES. So, 0.00039 " mol of Koh will heact with 9.000394 mol of HEPES to form 0.0003qv mol K-salt of HEPES So, amount of eneets HEPES = (6.023079569–0.00039u) mot Total volume = NmL. so, concentration of the acid HEPES. Paciót –(6.923079561-0-0003%) V ml Vmt & concentration of the K-salt of HEPES Balth = 0.00039 V mol This is an acidic buffer. Hendelson-Havelbaleh equi O pH= pkart dog (salt → log, salt 1 = pupka P 0.0003qu mot - 7.58-7.56 A Given, pH=7:58 bka= 7.56 0.0230795190.00039) mart Vm

volo.023079569 - 000039V) - ".02 0.00039 N 01 02 30795690.000390 0.093079569 -0.00039 Nein-0-02 0.000390 0.023079569 -1= 0.00034 V 0.954992586 0.093079569 =1.954992586 0.000390 0.02307956 Do 00039 *6.954992586 So, N = 30.27038694 noleume of Kott solution required, N= 30.27038694 mL ' 630.3 mL Kot volume = 30.3 ml →