Q. how to calculate change in pH

Q: How do I calculate pH from hydrogen ion concentration?

Use \mathrm{pH} = -\log_{10}\left[\mathrm{H}^+\right] . Plug in the molar concentration of hydrogen ions to get pH.

Q: How do I get hydrogen ion concentration from pH?

Use \left[\mathrm{H}^+\right] = 10^{-\mathrm{pH}} . This gives the molar concentration of hydrogen ions from known pH.

Q: How do I calculate pH of strong acid solution?

For monoprotic strong acid that fully dissociates, \left[\mathrm{H}^+\right] = c_0 . So \mathrm{pH} = -\log_{10} c_0 , where c_0 is the acid molarity.

Q: How do I calculate pH of weak acid?

Use K_= \frac{\left[\mathrm{H}^+\right]\left[\mathrm{A}^-\right]}{\left[\mathrm{HA}\right]} . Solve the quadratic from an ICE table, or approximate \left[\mathrm{H}^+\right]\approx\sqrt{K_c_0} if K_a/c_0 \ll 1 .

Q: How do I calculate pH of buffer?

Use Henderson-Hasselbalch: \mathrm{pH} = \mathrm{p}K_+ \log_{10}\frac{\left[\mathrm{A}^-\right]}{\left[\mathrm{HA}\right]} . Use concentrations or moles after any dilution or addition.

Q: How do activities and ionic strength affect pH calculations?

Use activity a_{\mathrm{H}^+}=\gamma_{\mathrm{H}^+}\left[\mathrm{H}^+\right] . Then \mathrm{pH} = -\log_{10}\left(\gamma_{\mathrm{H}^+}\left[\mathrm{H}^+\right]\right) . Estimate \gamm with Debye-Hückel or extended models at higher ionic strength.

Q: How do I calculate pH for polyprotic acids?

Use successive equilibrium constants K_{a1}, K_{a2}, \dots . Often the first dissociation dominates if K_{a1} \gg K_{a2} . Otherwise solve mass-balance and charge-balance with all equilibrior use stepwise approximations.

Answer

pH is defined by \( \mathrm{pH} = -\log_{10}[H^{+}]\). If the proton concentration changes from \([H^{+}]_{1}\) to \([H^{+}]_{2}\), the change in pH is
\[
\Delta \mathrm{pH} = \mathrm{pH}_{2}-\mathrm{pH}_{1} = -\log_{10}[H^{+}]_{2} + \log_{10}[H^{+}]_{1}
= -\log_{10}\!\left(\frac{[H^{+}]_{2}}{[H^{+}]_{1}}\right).
\]
For example, if \([H^{+}]\) decreases by a factor of 10 (i.e. \([H^{+}]_{2}=[H^{+}]_{1}/10\)), then
\[
\Delta \mathrm{pH} = -\log_{10}\!\left(\tfrac{1}{10}\right)=+1,
\]
so pH increases by 1.

Detailed Explanation

Definition and key formula. The pH of a solution is defined from the hydrogen ion concentration. Use the formula

\[ \mathrm{pH} = – \log_{10} \left( [\mathrm{H}^+] \right) \]

General approach to calculate a change in pH. Decide which case applies. Two common cases follow. For each case list the separate steps to perform, then carry out arithmetic.

Case 1 strong acid or strong base added to a nonbuffered solution. Steps to follow separately.

Step 1 Find initial moles of hydrogen ion and hydroxide ion. Multiply initial concentration by initial volume in liters.

Step 2 Find moles of acid or base added. Multiply concentration of titrant by volume added in liters.

Step 3 Perform neutralization. Subtract the smaller mole amount from the larger. If moles of H plus remain, the solution is acidic. If moles of OH minus remain, the solution is basic. If completely neutralized, the solution is neutral or determined by any remaining salt hydrolysis.

Step 4 Compute the final concentration of the excess species. Divide remaining moles by total volume after addition.

Step 5 Convert final concentration to pH. If excess H plus use pH equals negative log. If excess OH minus compute pOH equals negative log of [OH minus] then pH equals 14 minus pOH, assuming standard water autoionization constant at 25 degrees Celsius.

Step 6 Find the change in pH. Subtract initial pH from final pH to obtain delta pH.

Worked example for Case 1. Start with 50.00 milliliters of 0.1000 molar HCl. Add 10.00 milliliters of 0.01000 molar NaOH. Compute the change in pH.

Step A Convert volumes to liters and compute moles.

\[ n_{\mathrm{H}^+,\mathrm{initial}} = 0.05000\ \mathrm{L} \times 0.1000\ \mathrm{mol\ L^{-1}} = 0.005000\ \mathrm{mol} \]

\[ n_{\mathrm{OH}^-,\mathrm{added}} = 0.01000\ \mathrm{L} \times 0.01000\ \mathrm{mol\ L^{-1}} = 0.0001000\ \mathrm{mol} \]

Step B Neutralization. Subtract the smaller mole amount from the larger. Here acid in excess.

\[ n_{\mathrm{H}^+,\mathrm{remaining}} = 0.005000\ \mathrm{mol} – 0.0001000\ \mathrm{mol} = 0.004900\ \mathrm{mol} \]

Step C Total volume after addition.

\[ V_{\mathrm{total}} = 0.05000\ \mathrm{L} + 0.01000\ \mathrm{L} = 0.06000\ \mathrm{L} \]

Step D Final hydrogen ion concentration.

\[ [\mathrm{H}^+]_{\mathrm{final}} = \dfrac{0.004900\ \mathrm{mol}}{0.06000\ \mathrm{L}} = 0.0816667\ \mathrm{mol\ L^{-1}} \]

Step E Final pH.

\[ \mathrm{pH}_{\mathrm{final}} = – \log_{10} \left( 0.0816667 \right) \approx 1.088 \]

Step F Initial pH. For 0.1000 molar HCl the initial hydrogen ion concentration equals 0.1000 molar so

\[ \mathrm{pH}_{\mathrm{initial}} = – \log_{10} \left( 0.1000 \right) = 1.000 \]

Step G Change in pH.

\[ \Delta \mathrm{pH} = \mathrm{pH}_{\mathrm{final}} – \mathrm{pH}_{\mathrm{initial}} \approx 1.088 – 1.000 = 0.088 \]

Case 2 addition to a buffer. Use the Henderson Haasselbalch relation. Steps to follow separately.

Step 1 Identify the buffer acid base pair and the pKa of the weak acid. Step 2 Convert all volumes to liters and compute initial moles of weak acid HA and conjugate base A minus. Step 3 Add moles of strong acid or base. If adding strong acid, convert some A minus to HA. If adding strong base, convert some HA to A minus. Step 4 Compute new moles of HA and A minus. Step 5 Compute new pH with the Henderson Haasselbalch equation.

Henderson Haasselbalch formula in molar amounts reads

\[ \mathrm{pH} = \mathrm{p}K_a + \log_{10} \left( \dfrac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right) \]

When working with equal dilution volumes you may replace concentrations by moles divided by the same volume and the ratio reduces to a ratio of moles. Thus after addition the new pH may be computed as

\[ \mathrm{pH}_{\mathrm{final}} = \mathrm{p}K_a + \log_{10} \left( \dfrac{n_{\mathrm{A}^- , \mathrm{initial}} + n_{\mathrm{base\ added}} – n_{\mathrm{acid\ added\ neutralized}}}{n_{\mathrm{HA} , \mathrm{initial}} + n_{\mathrm{acid\ added}} – n_{\mathrm{base\ added\ neutralized}}} \right) \]

Practical simplified formula for a small addition of strong acid or base to a buffer. If you add a small amount of base, denote n_B the moles of base added and n_{A^-} and n_{HA} the initial buffer moles. Then

\[ \Delta \mathrm{pH} = \log_{10} \left( \dfrac{n_{\mathrm{A}^-} + n_B}{n_{\mathrm{HA}} – n_B} \right) \]

The sign will be reversed if you add acid instead of base. This expression follows from taking the difference of two Henderson Haasselbalch expressions for final and initial pH.

Worked example for Case 2. Start with a buffer containing 0.100 mol of HA and 0.100 mol of A minus. The pKa equals 4.76. Add 0.00100 mol of strong base. Compute the change in pH.

Step A New moles after addition of base.

\[ n_{\mathrm{A}^-,\mathrm{final}} = 0.100 + 0.00100 = 0.10100\ \mathrm{mol} \]

\[ n_{\mathrm{HA},\mathrm{final}} = 0.100 – 0.00100 = 0.09900\ \mathrm{mol} \]

Step B Final pH using Henderson Haasselbalch.

\[ \mathrm{pH}_{\mathrm{final}} = 4.76 + \log_{10} \left( \dfrac{0.10100}{0.09900} \right) = 4.76 + \log_{10} \left( 1.020202 \right) \approx 4.76 + 0.00873 = 4.7687 \]

Step C Initial pH with equal moles of A minus and HA. The ratio equals unity so initial pH equals pKa.

\[ \mathrm{pH}_{\mathrm{initial}} = 4.76 \]

Step D Change in pH.

\[ \Delta \mathrm{pH} = 4.7687 – 4.76 = 0.0087 \]

Summary and practical notes. For nonbuffered solutions perform mole bookkeeping and neutralization, then compute final concentration and apply pH definition. For buffer solutions use the Henderson Haasselbalch relation, update moles of HA and A minus after addition, then compute the new pH. For very small additions to a buffer a logarithmic approximation for delta pH using the ratio of final to initial amounts works well. Remember that temperature affects the water autoionization constant and the numeric relation pH plus pOH equals 14 is an approximation that is best at 25 degrees Celsius.

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Chemistry FAQs

How do I calculate pH from hydrogen ion concentration?

Use \( \mathrm{pH} = -\log_{10}\left[\mathrm{H}^+\right] \). Plug in the molar concentration of hydrogen ions to get pH.

How do I get hydrogen ion concentration from pH?

Use \( \left[\mathrm{H}^+\right] = 10^{-\mathrm{pH}} \). This gives the molar concentration of hydrogen ions from known pH.

How do I calculate pH of strong acid solution?

For monoprotic strong acid that fully dissociates, \( \left[\mathrm{H}^+\right] = c_0 \). So \( \mathrm{pH} = -\log_{10} c_0 \), where \( c_0 \) is the acid molarity.

How do I calculate pH of weak acid?

Use \( K_= \frac{\left[\mathrm{H}^+\right]\left[\mathrm{A}^-\right]}{\left[\mathrm{HA}\right]} \). Solve the quadratic from an ICE table, or approximate \( \left[\mathrm{H}^+\right]\approx\sqrt{K_c_0} \) if \( K_a/c_0 \ll 1 \).

How do I calculate pH of buffer?

Use Henderson-Hasselbalch: \( \mathrm{pH} = \mathrm{p}K_+ \log_{10}\frac{\left[\mathrm{A}^-\right]}{\left[\mathrm{HA}\right]} \). Use concentrations or moles after any dilution or addition.

How do I find the pH change when adding strong acid or base to buffer?

Subtract or add moles of added strong acid/base to the buffer species. Compute new moles, convert to concentrations, then apply Henderson-Hasselbalch with the updated ratio.

How do I calculate pH after mixing solutions of acids or bases?

Conserve moles. Compute total moles of acidic and basic species, divide by total volume for concentrations. For weak species set up ICE and solve equilibrium equations to get \( \left[\mathrm{H}^+\right] \).

How do activities and ionic strength affect pH calculations?

Use activity \( a_{\mathrm{H}^+}=\gamma_{\mathrm{H}^+}\left[\mathrm{H}^+\right] \). Then \( \mathrm{pH} = -\log_{10}\left(\gamma_{\mathrm{H}^+}\left[\mathrm{H}^+\right]\right) \). Estimate \( \gamm\) with Debye-Hückel or extended models at higher ionic strength.

How do I calculate pH for polyprotic acids?

Use successive equilibrium constants \( K_{a1}, K_{a2}, \dots \). Often the first dissociation dominates if \( K_{a1} \gg K_{a2} \). Otherwise solve mass-balance and charge-balance with all equilibrior use stepwise approximations.

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