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3 years ago

Calculate the pH value in each of the following solutions, given their [H3O+] concentrations.

Chemistry

1 answer:

FrozenT [24]3 years ago

4 0

Answer:

See Explanations ...

Explanation:

In general, pH is a 'p-factor' expression which, simply put, is a way to express very small numbers (i.e.; exponential data with 10⁻ⁿ value ranges) in a more convenient form. That is, by definition, pX = -log(X) where X is the data value of interest. In practical terms, p-factor analysis can be applied to a number of physical & chemical measurements such as ...

pH => measure of acidity of solution = -log[H₃O⁺]

pOH => measure of alkalinity of solution = -log[OH⁻]

pKa => measure of weak acid ionization in aqueous solution = -log(Ka)

pKb => measure of weak base ionization in aqueous solution = -log(Kb)

pKsp => measure of salt ionization in aqueous solution = -log(Ksp)

Such can be applied to ranges of small-number values defining other chemical and physical properties.

For this problem:

Gastric Juice: [H₃O⁺] = 1.6 x 10⁻²M => pH = -log(1.6 x 10⁻²) = -(-1.80) = 1.8

Cow's Milk: [H₃O⁺] = 2.5 x 10⁻⁷M => pH = -log(2.5 x 10⁻⁷) = -(-6.60) = 6.60

Tomato Juice: [H₃O⁺] = 5.0 x 10⁻⁵M => pH = -log(5.0 x 10⁻⁵) = -(-4.30) = 4.30

Other Applications:

Given:

[OH⁻] = 6.30 x 10⁻¹³M => pOH = -log(6.30 x 10⁻¹³) = -(-12.2) = 12.2

Ka = 4.5 x 10⁻⁵ => pKa = -log(4.5 x 10⁻⁵) = -(-4.35) = 4.35

Kb = 8.2 x 10⁻⁶ => pKb = -log(8.2 x 10⁻⁶) = -(-5.09) = 5.09

Ksp = 5.5 x 10⁻¹⁰ => pKsp = -log(-5.5 x 10⁻¹⁰) = -(-9.26) = 9.26

Note: The values for Ka, Kb & Ksp are typically provided in tables of weak acid ionization constants (Ka-values), weak base ionization constants (Kb-values) or solubility product constants of salts (Ksp-values).

Hope this helps, Doc :-)

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morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium

Free_Kalibri [48]

Moles of potassium permanganate = 0.0008

<h3>Further explanation </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002$

From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

$\tt \dfrac{2}{5}\times 0.002=0.0008$

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3 years ago

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How many significant gigures are in each mesasurments 303,000mm

Triss [41]

3 Sig FIgs in 303,000 mm
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3 years ago

If a reaction has an equilibrium constant just greater than 1 what type of reaction is it?

SashulF [63]

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