Calculate the pH of a solution prepared by mixing 50 mL of 0.1 M H2NaPO4 and 150

Question

Calculate the pH of a solution prepared by mixing 50 mL of 0.1 M H2NaPO4 and 150 mL of 0.1 M HNa2PO4 (pKa 7.21) Below is indicated the composition of an aqueous solution commonly used to carry out in organello assays to analyzed translation in isolated mitochondria. The solution contains: 0.6 M Sorbitol 150 mM KCI 15 mM phosphate buffer pH 7.4 12.66 mM MgClz 4 mMATP 500 M GTP 0.3% (w/v) BSA 5 mM Phosphoenolpyruvate 5 mM alpha-Ketoglutarate 0.24 mg/mL amino acid mix (no Cysteine-Tyrosine) 66.7 HM Cysteine 0.002% (w/v) Tyrosine 10 g/mL Pyruvate kinase How would you prepare 50 ml of this solution if you have available? Sorbitol (MW 182.17), KCl (MW 74.55), H2NaP04 (MW 119.98) and HNa2PO4 (Mw 141.96) (H2P04.pka 7.21), MgCl2(MW 95.21), 0.2 M ATP, 10 mM GTP, BSA, 1 M Phosphoenolpyruvate, 1 M alpha-Ketoglurate, 5 g/L amino acid mix, Cysteine (MW 121.16), Tyrosine and 1 mg/mL Pyruvate kinase.

Explanation / Answer

Formula: Molarity = no. of mmol/volume in mL

For NaH2PO4: 0.1 mmol/mL = no. of mmol/50 mL

i.e. The no. of mmol of NaH2PO4 = 0.1 mmol/mL * 50 mL = 5 mmol

For Na2HPO4: 0.1 mmol/mL = no. of mmol/150 mL

i.e. The no. of mmol of Na2HPO4 = 0.1 mmol/mL * 150 mL = 15 mmol

According to Henderson-Hasselbulch equation: pH = pKa2 + Log(nNa2HPO4/nNaH2PO4)

i.e. pH = 7.21 + Log(15/5)

i.e. pH = 7.21 + Log3

i.e. pH = 7.21 + 0.4771

i.e. pH = 7.687

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