What would you like to do?

# How would you Calculate the H3O ion concentration in a solution with pH3 and a solution with pH8?

# The pH of a solution is 6.0 what is the concentration in molL of hydronium ion in this solution and calculate the hydroxide ion concentration OH in molLM?

Important Notice: pH = negative value of the log 10 of the hyronium concentration , which is very low, mostly

# What solutions has the highest concentration of hydrogen ions pH10 pH8 pH6 or pH4?

Answer is pH = 4. Lower the pH, greater is the concentration of H + ions.

# If a solution has a pH of 7.5 what would it's new pH be if the concentration of H3O ions in the solution were increased by 100 times?

\npH=-log[H+] Where H+ and H3O is interchangeable.\nso with a pH=7.5 that means you have a [H+]= 10^(-7.5)\nIf you increase the hydrogen ion concentration by 100 you get\n.
…\n[H+]=100*10^(-7.5)=10^(-5.5)\nwhich gives you:\npH=5.5

# How do you calculate the H3O ion concentration in a solution with pH3 and ina solution with pH8?

pH is defined as the negative logarithm of the Hydronium ion concentration. Therefore, the concentration of Hydronium in a pH 3 solution is found in the following way. -Log(x)…=3, x=1*10^-3 M. If you are unfamiliar with the algebra of solving this equation you should review basic log rules. The log rule used to solve this problem is the following. Take the base of the log, 10, and raise it to each side of the equation. so, 10^(log(x))=10^-3 this reduces to x=10^-3.

# The concentration of an acid or base refers to the presence of H3O plus ions in solution?

False-apex

# A ph5 solution would have a corresponding hydronium ion concentration of?

[H 3 O + ] = 10 -pH = 10 -(5.0) = 1.0*10 -5 mol/L

# If a solution has a pH of 7.5 what would its new pH be if the concentration of h3O plus ions in the solution were increased by 100 times?

The new pH would be 5.5. pH = -log(H+) therefore 10 to the power of -5.5 = concentration of H3O+ ions. 10 to the power of -5.5 = 3.16x10 to the power of -8. multip…lied by 100 = 3.16x10 to the power of -6. -log(3.16x10 to the power of -6) = 5.5

# What is the H3O concentration in a solution with a pH of 2.34?

Since [H3O+] can be expressed as [H+]...the answer to your question is-----> [H+]=10^-pH.... which is 4.6x10^-3M

# What is the H3O plus ion concentration of a solution whose pH is 11?

pC of H 3 0 + at pH 11 is also 11, so concetration will be 10 -11 mol/l (anti-log of -11)

# What is the pH of an aqueous solution with the hydronium ion concentration H3O plus equals 2 x 10-14 M?

13.7 take the negative logarithm

# What would its new pH be if the concentration of H3O plus ions in the solution were increased by 100 times and why?

Going from right to left, each number on the ph scale represents a 10 fold increase in the concentration of H3O+ ions. For example, water has a ph of 7. Urine has 10 times mo…re H3O+ ions (than water) with a ph of 6. Vinegar has 10,000 more H3O+ ions (than water) with a ph of 3. An increase of 100 times would mean moving left on the ph scale by 2. If you started at say 7.5, subtract 2, to get 5.5.

# What is the pH of an aqueous solution with the hydronium ion concentration H3O 2 x 10-14 M?

To find the pH use this: -log(2 X 10^-14 M) = 13. 7 pH ( you could call it 14 )

Answered

In Chemistry

# What is the pH of an aqueous solution with the hydronium ion concentration H3O plus equals 2 x 10 14 M?

13.7

Answered

In Chemistry

# What is the H3O plus concentration of a solution with a pH of 9?

10 -9 = 1 x 10 -9 m ==

Answered

In Chemistry

# What is the pH of an aqueous solution with the hydronium ion concentration h3o plus equals 4 x 10-13m?

p = -log[H+] = 12.4

Answered

In Chemistry

# What is the pH of aqueous solution with the hydronium ion concentration H3O 210-14M?

2 x 10 -14 M pH = 13.7

Answered

In Chemistry

# The ph of lemon juice at 298 k is found to be 2.32 what is the concentration oh H3O plus ions in the solution?

If the PH of lemon juice at 298 k is found to be 2.32, theconcentration of H3O plus ions in the solution would be 0.5 M.