Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood is called the "bicarb number" and is used as a measure of blood \(\mathrm{pH}\) in hospital emergency rooms. A newly diagnosed diabetic patient is admitted to the emergency room with ketoacidosis and a bicarb number of \(10 .\) Calculate the blood \(\mathrm{pH} . K_{\mathrm{a}}\) for carbonic acid at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(7.9 \times 10^{-7}\)

Short Answer

Expert verified
The blood pH is 7.1.

Step by step solution

01

Understand the problem

We need to find the pH of the blood given the ratio (bicarb number) of \( \mathrm{HCO}_3^- \) to \( \mathrm{H}_2\mathrm{CO}_3 \) and the \( K_a \) of carbonic acid.
02

Apply the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is used to relate the ratio of the concentrations of the ionized form and the non-ionized form of an acid to the pH. It is given by: \[ \text{pH} = \text{pK}_a + \log \left( \frac{\left[\mathrm{HCO}_3^-\right]}{\left[\mathrm{H}_2\mathrm{CO}_3\right]} \right) \].
03

Calculate the pKa from Ka

First, calculate the \( \text{pK}_a \) from \( K_a \). Use the formula: \( \text{pK}_a = -\log( K_a ) \). Here, \( K_a = 7.9 \times 10^{-7} \).
04

Compute the pKa

Calculate \( \text{pK}_a \) using the expression: \[ \text{pK}_a = -\log(7.9 \times 10^{-7}) \]. This results in \( \text{pK}_a \approx 6.1 \).
05

Substitute values into the equation

Now substitute the \( \text{pK}_a \) and the bicarb number into the Henderson-Hasselbalch equation: \[ \text{pH} = 6.1 + \log(10) \].
06

Calculate the pH

Compute \( \log(10) \), which is 1, and complete the equation: \[ \text{pH} = 6.1 + 1 \] resulting in \( \text{pH} = 7.1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a crucial tool for calculating the pH of a solution, especially when dealing with weak acids and their conjugate bases. It simplifies the relationship between the pH of a solution, the pKa of the acid, and the concentrations of the acid and its conjugate base. This equation is represented as:\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]In the context of blood chemistry, the equation is used to ascertain the blood pH by considering the ratio of bicarbonate ions (\(\mathrm{HCO}_3^-\)) to carbonic acid (\(\mathrm{H}_2\mathrm{CO}_3\)). The pH level indicates whether the blood is more acidic or alkaline. Given the bicarb number, which is this ratio, medical professionals can quickly assess a patient's blood pH. It’s particularly useful in emergency situations like ketoacidosis, where time is of the essence, as it allows for rapid intervention to stabilize blood pH levels.
Carbonic acid
Carbonic acid plays a pivotal role in maintaining the acid-base balance in our blood. This weak acid forms when carbon dioxide dissolves in water, which is a common occurrence in our bodily functions.The dissociation of carbonic acid in blood is a key reaction: \[\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-\]In the bloodstream, this dissociation partly controls the concentration of hydrogen ions, thereby influencing the blood pH. Carbonic acid is an essential component of the bicarbonate buffer system, which ensures our blood remains within a tight pH range crucial for physiological processes.
Disturbances in levels of carbonic acid or bicarbonate ions can lead to health issues, emphasizing the importance of this component in not only maintaining pH balance but also in clinical diagnostics, especially for conditions like metabolic acidosis. Medical professionals often measure the ratio of bicarbonate to carbonic acid when assessing changes in blood pH.
Ketoacidosis
Ketoacidosis, particularly diabetic ketoacidosis (DKA), is a serious condition that arises when the body produces excess ketones, leading to an acidic blood environment. This is often seen in people with uncontrolled diabetes. During ketoacidosis, blood pH can drop to dangerously low levels as the accumulation of ketones increases acidity.
Symptoms of ketoacidosis include:
  • Frequent urination
  • Extreme thirst
  • High blood sugar levels
  • Nausea and vomiting
  • Fruity-scented breath
It occurs when there is not enough insulin to help the cells use glucose for energy, prompting the body to break down fat as an alternative energy source.
This breakdown process produces ketones, which are harmful in large amounts.
The body's buffer systems, such as the bicarbonate buffering system connected to carbonic acid, struggle to maintain equilibrium, further complicating conditions. In medical settings, the Henderson-Hasselbalch equation and the bicarb number are used to evaluate the severity and guide treatment of ketoacidosis. Immediate treatment often involves insulin therapy and rehydration to restore normal pH balance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each of the following, write the equilibrium-constant expression for \(K_{\mathrm{sp}}\) : (a) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (b) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) (c) \(\mathrm{BaCO}_{3}\) (d) \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\)

In an excess of \(\mathrm{NH}_{3}(a q), \mathrm{Cu}^{2+}\) ion forms a deep blue complex ion, \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), which has a formation constant \(K_{\mathrm{f}}=5.6 \times 10^{11} .\) Calculate the concentration of \(\mathrm{Cu}^{2+}\) in a solution prepared by adding \(5.0 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CuSO}_{4}\) to \(0.500 \mathrm{~L}\) of \(0.40 \mathrm{M} \mathrm{NH}_{3}\).

Assume that \(40.0 \mathrm{~mL}\) of a \(0.0250 \mathrm{M}\) solution of the protonated form of the amino acid valine \(\left(\mathrm{H}_{2} \mathrm{~A}^{+}\right)\) is titrated with \(0.100 \mathrm{M} \mathrm{NaOH}\). Calculate the \(\mathrm{pH}\) after addition of the following volumes of \(0.100 \mathrm{M} \mathrm{NaOH}\) : (a) \(10.0 \mathrm{~mL}\) (b) \(15.0 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL}\) $$ \begin{aligned} &\text { Valine cation }\left(\mathrm{H}_{2} \mathrm{~A}^{+}\right)\\\ &\begin{aligned} &K_{\mathrm{a} 1}=4.8 \times 10^{-3} \\ &K_{\mathrm{a} 2}=2.4 \times 10^{-10} \end{aligned} \end{aligned} $$

Use the following solubility data to calculate a value of \(K_{\mathrm{sp}}\) for each compound: (a) \(\mathrm{CdCO}_{3} ; 1.0 \times 10^{-6} \mathrm{M}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2} ; 1.06 \times 10^{-2} \mathrm{M}\) (c) \(\mathrm{PbBr}_{2} ; 4.34 \mathrm{~g} / \mathrm{L}\) (d) \(\mathrm{BaCrO}_{4} ; 2.8 \times 10^{-3} \mathrm{~g} / \mathrm{L}\)

Explain why the uptake of \(\mathrm{CO}_{3}{ }^{2-}\) by calcifying marine organisms causes more \(\mathrm{CO}_{2}\) to dissolve in the ocean than in pure water.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free