# pH calculator for acids, bases and buffers

> Calculate pH, pOH, [H⁺] and [OH⁻] from a concentration, a strong acid or base, a weak acid's Ka or pKa, or a Henderson–Hasselbalch buffer.

ఇంటరాక్టివ్ వెర్షన్: https://www.calcopenly.com/te/science/ph-calculator
విషయం: విజ్ఞాన శాస్త్ర కాలిక్యులేటర్లు

pH is the negative base-10 logarithm of the hydrogen-ion concentration, pH = −log₁₀[H⁺]. At 25 °C the ion product of water, Kw, is 1.0 × 10⁻¹⁴, so pH + pOH = 14. Strong acids and bases are treated as fully dissociated, with water's own ions included; weak acids are solved exactly from Ka with the quadratic x² + Ka·x − Ka·C = 0; buffers use the Henderson–Hasselbalch equation.

The default, 0.1 mol/L acetic acid with Ka = 1.8 × 10⁻⁵, gives pH 2.88 with 1.33 % of the acid ionised. The √(Ka·C) shortcut taught in class gives pH 2.872, because it overstates [H⁺] by 0.67 %.

Concentrations stand in for activities, which works best for dilute solutions. Results below pH 0 or above 14 are possible for concentrated acids and bases, but there the simple formula is only an estimate.

## ఇన్‌పుట్‌లు

- **Start from** (ఎంపికలు: [H⁺] concentration, [OH⁻] concentration, pH value, Strong acid concentration, Strong base concentration, Weak acid (Ka), Buffer (Henderson–Hasselbalch))
- **Concentration**
- **Concentration unit** (ఎంపికలు: µmol/L, mmol/L, mol/L)
- **H⁺ or OH⁻ released per formula unit**: 2 for Ba(OH)₂; H₂SO₄'s second proton is only partly dissociated
- **pH**
- **Acid strength given as** (ఎంపికలు: Ka, pKa)
- **Ka**
- **pKa**
- **Conjugate base [A⁻]**
- **Weak acid [HA]**

## ఫలితాలు

- pH — ప్రధాన ఫలితం
- pOH
- [H⁺] (mol/L)
- [OH⁻] (mol/L)
- Solution is
- pH with the √(Ka·C) shortcut
- Shortcut error in [H⁺]
- Fraction ionised

## సూత్రం

$$
\text{pH} = -\log_{10}[\mathrm{H^+}],\ \text{pH} + \text{pOH} = 14;\quad x^2 + K_a x - K_aC = 0;\quad \text{pH} = \text{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}
$$

## పరిష్కరించిన ఉదాహరణలు

### [H⁺] = 0.001 mol/L

- Start from: [H⁺] concentration
- Concentration: 0.001
- Concentration unit: mol/L
- **pH: 3.00**
- **pOH: 11.00**
- **Solution is: Acidic**
- తనిఖీ చేసిన మూలం: −log₁₀(10⁻³) = 3; pOH = 14 − 3

### pH 7 is neutral

- Start from: pH value
- pH: 7
- **[H⁺]: 1 × 10⁻⁷ mol/L**
- **[OH⁻]: 1 × 10⁻⁷ mol/L**
- **Solution is: Neutral**
- తనిఖీ చేసిన మూలం: 10⁻⁷ mol/L at 25 °C (Kw = 10⁻¹⁴)

### 0.01 M HCl

- Start from: Strong acid concentration
- Concentration: 0.01
- Concentration unit: mol/L
- H⁺ or OH⁻ released per formula unit: 1
- **pH: 2.00**
- తనిఖీ చేసిన మూలం: Python 3.8 decimal: [H⁺] = (C + √(C² + 4Kw))/2 = 0.010000000001 ⇒ pH 2.0000

### Very dilute HCl (1e-8 M) is still acidic

- Start from: Strong acid concentration
- Concentration: 1e-8
- Concentration unit: mol/L
- H⁺ or OH⁻ released per formula unit: 1
- **pH: 6.98**
- **[H⁺]: 1.05125 × 10⁻⁷ mol/L**
- తనిఖీ చేసిన మూలం: Python 3.8 decimal: [H⁺] = (10⁻⁸ + √(10⁻¹⁶ + 4×10⁻¹⁴))/2 = 1.0512492×10⁻⁷ ⇒ pH 6.97829 (not 8)

### 0.1 M acetic acid, Ka = 1.8e-5

- Start from: Weak acid (Ka)
- Concentration: 0.1
- Concentration unit: mol/L
- Acid strength given as: Ka
- Ka: 1.8e-5
- **pH: 2.88**
- **pH with the √(Ka·C) shortcut: 2.872**
- **Shortcut error in [H⁺]: 0.67%**
- తనిఖీ చేసిన మూలం: Python 3.8 decimal: x = (−Ka + √(Ka² + 4KaC))/2 = 1.33267×10⁻³ ⇒ pH 2.87528; √(KaC) ⇒ 2.87236

### Buffer with [A⁻] = 2[HA]

- Start from: Buffer (Henderson–Hasselbalch)
- pKa: 4.76
- Conjugate base [A⁻]: 0.2 mol/L
- Weak acid [HA]: 0.1 mol/L
- **pH: 5.06**
- తనిఖీ చేసిన మూలం: Python 3.8 decimal: 4.76 + log₁₀2 = 5.0610300

## ప్రశ్నలు

### How do you calculate pH from concentration?

For a strong acid, [H⁺] equals the acid concentration times the protons each unit releases, and pH = −log₁₀[H⁺]: 0.01 M HCl has pH 2.00. For a strong base, take pOH = −log₁₀[OH⁻] and subtract it from 14: 0.005 M Ba(OH)₂ releases 0.01 M OH⁻, so pOH is 2 and pH is 12. Weak acids only partly ionise, so they also need Ka.

### Why is the pH of pure water 7?

Water ionises slightly, H₂O ⇌ H⁺ + OH⁻, and at 25 °C the product [H⁺][OH⁻] equals Kw = 1.0 × 10⁻¹⁴. In pure water the two concentrations are equal, so each is 10⁻⁷ mol/L and the pH is 7. Kw grows with temperature: at 50 °C pKw is 13.26, so neutral water has pH 6.63. This calculator uses the 25 °C value.

### What is the difference between Ka and pKa?

Ka is the acid dissociation constant, [H⁺][A⁻]/[HA] at equilibrium, and pKa = −log₁₀Ka. A lower pKa means a stronger acid, and each unit of pKa is a factor of 10 in Ka. Acetic acid has pKa 4.76, so Ka = 1.74 × 10⁻⁵ (often rounded to 1.8 × 10⁻⁵); hydrofluoric acid, Ka = 6.8 × 10⁻⁴, has pKa 3.17.

### When is the Henderson–Hasselbalch equation accurate?

pH = pKa + log₁₀([A⁻]/[HA]) holds when the acid and its conjugate base are both far more concentrated than Ka and [H⁺], and their ratio lies between 0.1 and 10, which keeps the pH within one unit of pKa. Outside that ratio the buffer has little capacity left, and the calculator warns you. Equal concentrations give pH = pKa, so a 1:1 acetate buffer sits at 4.76.

### Can pH be negative?

Yes. A 2 mol/L solution of a strong acid such as HCl has [H⁺] of about 2 mol/L, and −log₁₀2 gives pH −0.30. At concentrations like this, ion activities differ markedly from concentrations, so a pH electrode reads a different value. Treat any calculated pH below 0 or above 14 as an estimate rather than a measurement.

### “pH calculator for acids, bases and buffers” ఎంత కచ్చితమైనది?

కచ్చితత్వం మీ ఇన్‌పుట్‌లు, పద్ధతిలోని ఊహలపై ఆధారపడి ఉంటుంది. దశాంశ గణన 50 సార్థక అంకెలను ఉపయోగిస్తుంది. అంచనాలు, సంఖ్యా పద్ధతులు, మూల డేటా తక్కువ కచ్చితత్వంతో ఉండవచ్చు; ప్రదర్శనలో విలువలను రౌండ్ చేయడం ఈ పరిమితులను తొలగించదు. స్వతంత్ర మూలాల పరిష్కారాలతో తనిఖీ చేసిన ఉదాహరణలు: 8. ఉదాహరణకు, “[H⁺] = 0.001 mol/L”ను −log₁₀(10⁻³) = 3; pOH = 14 − 3తో తనిఖీ చేశారు.

### ఈ పద్ధతికి మూలం ఏమిటి?

OpenStax Chemistry 2e, §14.2 pH and pOH; §14.3 Relative strengths of acids and bases; §14.6 Buffers; IUPAC Gold Book — pH; ionic product of water (Kw = 1.0 × 10⁻¹⁴ at 25 °C).

## మూలాలు

- [OpenStax Chemistry 2e, §14.2 pH and pOH; §14.3 Relative strengths of acids and bases; §14.6 Buffers](https://openstax.org/books/chemistry-2e/pages/14-2-ph-and-poh)
- [IUPAC Gold Book — pH; ionic product of water (Kw = 1.0 × 10⁻¹⁴ at 25 °C)](https://goldbook.iupac.org/terms/view/P04524)