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modified true/false: if the statement is true, write true in the blank.…

Question

modified true/false: if the statement is true, write true in the blank. if it is false, replace the italicized word(s) with the word(s) that will make the statement true.

  1. a solution in which $\ce{h3o+} = \ce{oh-}$ is acidic.
  2. a solution with a ph of 2.3 is basic.
  3. litmus, bromocresol purple, and phenolphthalein are acid - base indicators.

short answer: write the correct answer in the blank.

  1. what is defined as the negative logarithm of hydronium concentration?
  2. what unit equals the mole amount of acid that will supply 1 mol of $\ce{h+}$ or the amount of base that will supply 1 mol of $\ce{oh-}$?
  3. what type of device determines ph by measuring the electric current through a solution?

problems: solve the problem; write the answer in the blank using proper significant figures and units.

  1. what is the normality of a 0.10 m solution of the triprotic acid phosphoric acid?
  2. what is the ph of a solution in which $\ce{h3o+} = 1.0 \times 10^{-5}\ce{m}$?
  3. the ph of a solution is 10.00; what is $\ce{h3o+}$ in this solution?
  4. the poh of a solution is 2.0; what is its ph?

Explanation:

Problem 7:

Step1: Recall Normality Formula

Normality (\(N\)) for an acid is given by \(N = M \times n\), where \(M\) is molarity and \(n\) is the number of \(H^+\) ions donated per molecule. For phosphoric acid (\(H_3PO_4\)), it's triprotic, so \(n = 3\).

Step2: Substitute Values

Given \(M = 0.10\) M and \(n = 3\), calculate \(N\): \(N = 0.10 \, M \times 3 = 0.30 \, N\).

Step1: Recall pH Formula

The formula for pH is \(pH = -\log[H_3O^+]\).

Step2: Substitute \([H_3O^+]\)

Given \([H_3O^+] = 1.0 \times 10^{-5} \, M\), substitute into the formula: \(pH = -\log(1.0 \times 10^{-5})\). Since \(\log(1.0 \times 10^{-5}) = -5\), then \(pH = -(-5) = 5\).

Step1: Recall pH to \([H_3O^+]\) Formula

The formula relating pH and \([H_3O^+]\) is \([H_3O^+] = 10^{-pH}\).

Step2: Substitute pH Value

Given \(pH = 10.00\), substitute into the formula: \([H_3O^+] = 10^{-10.00} = 1.0 \times 10^{-10} \, M\).

Answer:

\(0.30 \, N\)

Problem 8: