# What is the h+ if the ph of a solution is 1.65?

What is the h+ if the ph of a solution is 1.65?

## This Post Has 6 Comments

1. kcopeland210 says:

The  $H+\\$ concentration is $1.41 * 10^{-5}\\$

Explanation:

The pH of any solution is given by the equation

$pH = -log [H+]$

Where $H+\\$ is the hydrogen ion concentration

Substituting the given values in above equation, we get -

$4.85 = -log[H+]\\H += 1.41 * 10^{-5}\\$

2. annjetero2oy23ay says:

[H⁺] = 0.920 MpH = 0.036[OH⁻] = 1.086x10⁻¹⁴ M

Explanation:

To calculate [H⁺] first we calculate the total number of H⁺ ions.

H⁺ from HCl ⇒ 49.3 mL * 1.19 gSolution/mL * 38 gHCl / 100 gSolution = 22.29 g HCl

22.29 g HCl ÷ 36.45 g/mol = 0.612 mol HCl = 0.612 mol H⁺

H⁺ from HNO₃ ⇒ 19.5 mL * 1.42 gSolution/mL * 70 gHCl / 100 gSolution = 19.38 g HNO₃

19.38 g HNO₃ ÷ 63 g/mol = 0.308 mol HNO₃ = 0.308 mol H⁺

Total H⁺ moles = 0.612 + 0.308 = 0.920 mol H⁺

Final Volume = 1.00 L

[H⁺] = 0.920 mol / 1.0 L = 0.920 M

Now we calculate pH:

pH = -log [H⁺]pH = -log(0.920) = 0.036

To calculate [OH⁻], we calculate pOH:

pOH = 14 - pHpOH = 14 - 0.036 = 13.964

pOH = 13.964 = -log[OH⁻]

$10^{-13.964}$ = [OH⁻] = 1.086x10⁻¹⁴ M

3. jjaheimhicks3419 says:

the concentration of H+ = 3 ×  $10^{-3}$   M and the pH = 2.6 of the solution

Explanation:

Based on reduction potentials, hydrogen is a better reducing agent than copper, therefore copper($Cu^{2+}$) is the cathode and hydrogen ($H_{2}$) is the anode.

Cathode reaction (reduction): $Cu^{2+}(aq) + 2e^{-}$  ⇒   Cu(s)

Anode reaction (oxidation) :  $H_{2}(g)$     ⇒    $2H^{+}(aq) + 2e^{-}$

net reaction:   $H_{2}(g) +$    $Cu^{2+}(aq)$   ⇒  $2H^{+}(aq) + Cu(s)$

$E_{0}cell = E_{0}cathode - E_{0}anode$

E cathode = 0.337 v

$E_{0}cell = + 0.337 - 0 = 0.337$

Q(reaction quotient) = $\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }$

for 2 electrons $Cu^{2+} = 1.00M$ but $H^{+}$ is unknown. we solve this using hernst equation.

$E = E^{0} -\frac{0.0257}{n}ln\frac{[H^{+}]^{2} }{[Cu^{2+}]P_{H2} }$

$0.490 = 0.337 -\frac{0.0257}{2}ln\frac{[H^{+}]^{2} }{[1][1]}$

$ln{[H^{+}]^{2} } = -11.9$

$2ln{[H^{+}] } = -11.9$

$ln{[H^{+}] } = -5.95$

$[H^{+}] = 3* 10^{-3} M$

pH = 2.6

4. SunsetPrincess says:

The pH of a solution is measure of the acidity of a certain solution based from the concentration of the hydrogen ions. It is associated with the hydrogen ion dissolved in the solution. It is expressed as pH = -log [H+]. We calculate the concentration of the hydrogen ions from this expression.

pH = -log [H+]
1.65 = -log [H+]
antilog [- 1.65] = [H+]
[H+] = 10^-1.65
[H+] = 0.0224 M

5. aaronpmoore1010 says:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

6. huwoman says:

The pH of a solution is a value used to measure the acidity of a solution and also a measure of the hydrogen ion present in the solution. It is logarithmically related to the hydrogen ion concentration. It is expressed as:

pH = -log [H+]
4.85 = -log[H+]
[H+] = 1.41x10^-5 M