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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.

నవీకరించబడింది తనిఖీ చేసిన ఉదాహరణలు: 8

= 0.000018
ప్రయత్నించండి
pH
pH: 2.88
దశాంశ స్థానాలు: 2; దగ్గరి విలువ; సమదూరంలో సరి అంకె వైపు
pOH
11.12
[H⁺]
0.00133267mol/L
[OH⁻]
7.50373 × 10⁻¹²mol/L
Solution is
Acidic
pH with the √(Ka·C) shortcut
2.872
Shortcut error in [H⁺]
0.67%
Fraction ionised
1.33%
Uses Kw = 1.0 × 10⁻¹⁴ (25 °C) and concentrations in place of activities.

pH 2.88 (acidic): [H⁺] = 0.001333 mol/L and [OH⁻] = 7.504 × 10⁻¹² mol/L; the acid is 1.33% ionised, and the √(Ka·C) shortcut would be off by 0.67%.

pH scale

Strongly acidicAcidicNear neutralBasicStrongly basic03681114neutral, pure waterpH 2.88
గణన చేసే విధానం S

  1. Equilibrium HA ⇌ H⁺ + A⁻

    Ka=x2C−x⇒x2+Kax−KaC=0,Ka=0.000018, C=0.1K_a = \frac{x^2}{C - x} \Rightarrow x^2 + K_a x - K_aC = 0,\quad K_a = 0.000018,\ C = 0.1

  2. Exact root

    x=−Ka+Ka2+4KaC2=0.00133267 mol/Lx = \frac{-K_a + \sqrt{K_a^2 + 4K_aC}}{2} = 0.00133267\ \mathrm{mol/L}

    Water's own ionisation is neglected — fine while [H⁺] ≫ 10⁻⁷ mol/L.

  3. Compare the shortcut

    KaC=0.00134164⇒pH≈2.8724 (0.673% high in [H+])\sqrt{K_aC} = 0.00134164 \Rightarrow \text{pH} \approx 2.8724\ (0.673\%\ \text{high in }[\mathrm{H^+}])

  4. pOH and [OH⁻] at 25 °C

    pH=2.87528,pOH=14−pH=11.1247,[OH−]=7.50373×10−12 mol/L\text{pH} = 2.87528,\quad \text{pOH} = 14 - \text{pH} = 11.1247,\quad [\mathrm{OH^-}] = 7.50373 \times 10^{-12}\ \mathrm{mol/L}

pH calculator for acids, bases and buffers గురించి

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.

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)

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).

ఈ కాలిక్యులేటర్ గురించి

pH=−log⁡10[H+], pH+pOH=14;x2+Kax−KaC=0;pH=pKa+log⁡10[A−][HA]\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}]}

మూలాలు

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

ఆధారాలతో సరిచూసినవి

ఈ కాలిక్యులేటర్‌లో స్వతంత్ర ఆధారాల నుంచి సమాధానాలు పొందిన 8 సాధించిన ఉదాహరణలు ఉన్నాయి. ఇవి పరీక్షల సమూహంలో అమలవుతాయి; మీరు ఇక్కడ కూడా వాటిని అమలు చేయవచ్చు.

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