The pH of a buffer system is dependant upon the temperature. Therefore, it is important to measure the pH at the temperature of the assay.

Note: this calculator assumes the pH is close to neutral. A large deviation from neutrality may violate key assumptions.

The ionic strength of solution is calculated by the following equation:

$$\frac{1}{2} \sum_{i=1}^n M_i \times z^2_i$$

It is important to remember that a buffer is composed of weak acid and conjugate base. The concentration and charge of these species is dependant upon the pKa of the specific buffer and the pH and temperature of the system. If these values are known, the percent ionization can be easily calculated from the Henderson–Hasselbalch equation:

$$\text{pH} = \text{pKa} + \log\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}$$

Reach Devices has an excellent pKa calculator for many biological buffers.

As an example, consider a 100 mM buffer of Potassium MES at 37°C, pH 7.2. The pKa of MES is 6.01:

$$7.2 = 6.01 + \log\frac{[\text{MES}^{-1}]}{[\text{H}\cdot\text{MES}]}$$

Solving the Henderson–Hasselbalch equation shows that ~94% of the MES will be ionized. Therefore, the ionic strength is calculated:

$$\frac{1}{2} \times (94 \text{mM} \times (-1)^2 + 100 \text{mM} \times (+1)^2) = 97\text{mM}$$

Be sure to perform this calculation for all ionic species in your system

The free magnesium concentration is an important component of the calculation

Ideally, this value will be empirically determined under the specific experimental conditions of the CK Clamp.

As an alternative, MaxChelator can be used to estimate the free magnesium of the system.

The total concentration of PO₄^3- in solution.

NOTE:

The pKa₂ of 10mM phosphoric acid is ~7. If the pH of your buffer is close to this value, the Ionic Strength calculation should account for the concentration of monobasic and dibasic anions.

For example, consider a buffered solution at pH 7 containing 10 mM KH₂PO₄. Solving the Henderson–Hasselbalch equation shows the ions will be 45% monobasic and 50% dibasic in the solution:

$$\text{pH} = \text{pKa} + \log\frac{\text{H}\text{PO}_4^{2-}}{\text{H}_2\text{PO}_4^{1-}}$$