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4 years ago

If the [H+] in a solution is 1 × 10–1 mol/L, what is the [OH–]? Show your work.

1 answer:

3 0

Hello!

The basic equations to solve this is
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
------------------------------------------------------------------------------------------------------
Find pH

pH = -log(1 * 10^-1)
pH = 1
------------------------------------------------------------------------------------------------------
Find pOH
1 + pOH = 14

pOH = 13
------------------------------------------------------------------------------------------------------
Find OH-
[OH-] = 10^(-pOH)
[OH-] = 1 * 10^-13mo/L

The answer is $[OH-] = 1 * 10^{-13} mol/L$

Hope this helps!

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Question 1 (Multiple Choice Worth 2 points)

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2 years ago

Calculate the amount in grams of Na2CO3 needed to react with HCL to produce 120g NaCl

worty [1.4K]

Let's start off with the balanced chemical equation:

$Na_2CO_{3_{(s)}} + \textbf2HCl_{_{(aq)}}\rightarrow CO_{2_{(g)}} + H_2O_{_{(l)}} + \textbf2NaCl_{_{(aq)}}$

Assuming the sodium carbonate is the limiting reagent, look at the coefficients of sodium carbonate and sodium chloride and use those as a ratio of sodium carbonate to sodium chloride: 1:2.

Since you have the required mass of NaCl, convert this to moles.

Assuming you know how to find the molar mass of NaCl:

$M_{NaCl} = 58.44g/mol$
$n = \frac{m}{M}$
$n_{NaCl} = \frac{120g}{58.44g/mol}$
$n_{NaCl} = 2.053mol$

Using the ratio, since 1 mole of sodium carbonate is required to produce 2 moles of sodium chloride, cross-multiply the ratios:

$1:2 = x:2.053mol$
$2x = 2.053mol$
$x = 1.027mol$

Therefore 1.027 moles of sodium carbonate is required to produce the required amount of sodium chloride. Convert to mass for your final answer:

$n_{Na_2CO_3} = 1.027mol$
$M_{Na_2CO_3} = 105.99g/mol$
$m = nM$
$m = (1.027mol)(105.99g/mol)$
$m_{Na_2CO_3} = 108.85g$

Hope this helps :)

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3 years ago

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