Phosphate Buffer pH Discrepancy Before an Enzyme Assay
You are preparing 1.00 L of 50.0 mM potassium phosphate buffer, pH 7.40, for an enzyme assay. The stock solutions are 0.200 M KH2PO4 (monobasic) and 0.200 M K2HPO4 (dibasic). At 25 °C the second pKa of phosphoric acid is 7.20. You combine 125 mL of each stock, dilute to 1.00 L with deionized water, and measure pH 7.28 with a glass electrode calibrated two hours earlier. A colleague suggests adding 1.0 M NaOH dropwise until the meter reads 7.40. The assay is sensitive to phosphate concentration and ionic strength and will run at 25 °C. The pH meter temperature probe is set to 25 °C. The morning calibration slope was 92%, but the technician accepted it because pH 7.00 buffer read 7.02. The buffer will be used within 4 h.
1. Calculate the expected pH of the 125 mL + 125 mL mixture and the stock volumes needed for pH 7.40. Is the measured 7.28 credible?
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Equal volumes of 0.200 M monobasic and dibasic stocks give equal concentrations of H2PO4- and HPO4^2-, so the Henderson-Hasselbalch ratio is 1 and the expected pH is pKa2 = 7.20 at 25 °C. The measured 7.28 is 0.08 pH units high, which is larger than the normal agreement expected from a properly calibrated electrode. The 92% slope is the stronger warning: a pH electrode should have a slope near 95-105% of theoretical. With a 92% slope, the meter can read close to the calibration point but deviate increasingly away from it; accepting the calibration because pH 7.00 read 7.02 does not validate the slope. Therefore 7.28 should not be trusted until the electrode is recalibrated with fresh pH 7.00 and 4.00 or 10.00 buffers and the slope is acceptable.
For pH 7.40, the required base/acid ratio is 10^(7.40-7.20) = 10^0.20 = 1.585. The total phosphate is 50.0 mmol in 1.00 L. Let acid volume be V_acid and base volume be V_base from 0.200 M stocks. V_base/V_acid = 1.585 and V_base + V_acid = 250.0 mL, because 250.0 mL × 0.200 M = 50.0 mmol. Solving gives V_acid = 96.7 mL and V_base = 153.3 mL. Dilute to 1.00 L. This buffer would be 50.0 mM in total phosphate and would have the intended ratio if the stocks and water are accurate.
2. Assuming the true pH of the equimolar buffer is 7.20, calculate how much 1.0 M NaOH would be needed to reach pH 7.40 in 1.00 L. Should you do that instead of remaking the buffer?
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The 1.00 L equimolar buffer contains 25.0 mmol H2PO4- and 25.0 mmol HPO4^2-. Adding x mmol OH- converts H2PO4- to HPO4^2-, so the final ratio is (25.0 + x)/(25.0 - x) = 1.585. Solving: 25.0 + x = 39.625 - 1.585x, so 2.585x = 14.625 and x = 5.66 mmol. That requires 5.66 mL of 1.0 M NaOH. The volume change is only about 0.57%, but the added Na+ raises ionic strength and the local pH during addition can be high unless it is stirred well.
I would not use the dropwise NaOH approach here, for three reasons. First, the pH meter is not trustworthy because of the 92% slope, so adding NaOH to a displayed 7.40 could leave the true pH significantly off. Second, the assay is stated to be sensitive to ionic strength; adding 5.66 mM Na+ changes the matrix even if the phosphate ratio is corrected. Third, remaking the buffer from the calculated stock volumes is fast and gives a defined composition. The correct sequence is to recalibrate the electrode, verify the pH of fresh pH 7.00 buffer, remake the buffer using 96.7 mL monobasic stock plus 153.3 mL dibasic stock, and check the final pH at 25 °C. If the remade buffer still reads outside 7.40 ± 0.05, investigate the stocks or electrode rather than blindly adding acid or base.